mm-3129 === CRANK.NET http://www.crank.net/harris.html === Subject: Quick integration question Real simple one here I'm sure, its just so long I've used calculus I'm having a hard time, I have a function I need to integrate which is C * LN( T - X) with respect to X, and for the life of me I cant remember how, can anyone help me with this (C & T) are constants === Subject: Re: Quick integration question > Real simple one here I'm sure, its just so long I've used calculus I'm > having a hard time, I have a function I need to integrate which is > C * LN( T - X) > with respect to X, and for the life of me I cant remember how, can > anyone help me with this (C & T) are constants Do a substitution and then integrate by parts. === Subject: Re: Quick integration question is this right...? (i will use this sign Û as an integral sign) ÛC*LN(T-X)dX I've made a substitution U = T-X ....then DU = -DX then DX = -DU = - C*Û Ln(U)dU (C is constant so it goes in front of the integral) ÛLn(U)dU is a integral that you can find in integral tables, and it is.... ÛLn(X)dX = X*Ln(X)-X+A so.... = - C*Û Ln(U)dU= - C * [U * Ln(U) - U + A] and finally = - C * [(T-X)* Ln(T-X) - (T-X) + A] where A is constant.. Is that solution right? I think it is....my humble opinion :) ÛC*Ln(T-X)dX = - C*[ (T-X)* Ln(T-X) - (T-X) + A] >> Real simple one here I'm sure, its just so long I've used calculus I'm >> having a hard time, I have a function I need to integrate which is >> C * LN( T - X) >> with respect to X, and for the life of me I cant remember how, can >> anyone help me with this (C & T) are constants > Do a substitution and then integrate by parts. > Dave === Subject: Re: Quick integration question > is this right...? (i will use this sign Û as an integral sign) > ÛC*LN(T-X)dX > I've made a substitution U = T-X ....then DU = -DX then DX = -DU > = - C*Û Ln(U)dU (C is constant so it goes in front of the integral) > ÛLn(U)dU is a integral that you can find in integral tables, and it is.... > ÛLn(X)dX = X*Ln(X)-X+A > so.... = - C*Û Ln(U)dU= - C * [U * Ln(U) - U + A] > and finally = - C * [(T-X)* Ln(T-X) - (T-X) + A] > where A is constant.. > Is that solution right? I think it is....my humble opinion :) > ÛC*Ln(T-X)dX = - C*[ (T-X)* Ln(T-X) - (T-X) + A] > Real simple one here I'm sure, its just so long I've used calculus I'm > having a hard time, I have a function I need to integrate which is > C * LN( T - X) > with respect to X, and for the life of me I cant remember how, can > anyone help me with this (C & T) are constants >> Do a substitution and then integrate by parts. >> Dave Looks good to me! === Subject: Re: Quick integration question C*(X-1)*Ln(T-X) === Subject: Re: Quick integration questiont: http://integrals.wolfram.com/index.jsp === Subject: math tutors hand book A great resource for math teachers, tutors, homeschoolers or studying for standardized test, or personal review! Check out : www.janzia.com you'll love it! === Subject: Re: COME BACK SATCHI ALL IS FORGIVEN, was: The Speed of Toilet Paper > it's http://www.odd-info.com. === Subject: help on a proof I have a question on how to prove E(sup_n(Y_n)) < infinity. Known: (1) Y_n >=0 is random variables and bounded for any n (n=1,2,...); (2) lim_{n->infinity} Y_n = 0; Question: can we prove: E(sup_n(Y_n)) < infinity? === Subject: Re: SF: Progress, double factorization proven The k's are determined as they are a finite set of integers. > : That is, with your target T odd and surrogate S odd, you are looking > : for solutions where ALL The k's are integers, which is a finite set. > You haven't adequately explained why this new set (these new sets in fact) > of integers is (are) easier to deal with than just looking at all the > integers that *might* divide T, which is after all also a finite set. > I gave the key equations that give the simple answer to those kinds of > questions: > with S = f_1*f_2 and T = g_1 * g_2 > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > so if there are, say, 4 integer factorizations of T, and 4 integer > factorizations of S, then you have 16 possible integers. > So even if T is your public key, from that perspective, a space of 16 > solutions would exist, which is a lot smaller than the set of all > itnegers that might divide your public key. Just because the set is smaller doesn't mean it's any easier to deal with. A = { n : n is a minimal counterexample to Goldbach's Conjecture } is a set with at most one element, and B = { n : n is a positive divisor of 4 } is a set with exactly 3 elements. However, no one knows what the set A is, or even if it's nonempty, whereas B is easy to calculate. (Goldbach's conjecture states that every even number >= 6 can be written as the sum of two (positive) prime numbers, so A would be the smallest even integer which cannot be written as the sum of two prime numbers, IF such a number exists.) --- Christopher Heckman > It's not only relevent that the set's finite, what's important is that you > have some sort of efficient way of going through that finite calculation. > All you did is rephrase the problem. > I've given a lot of detail in this thread and my latest thread. > Going to give more detail now in just a bit. > Oh yeah, as a sidenote, x=y=1 is a trivial case. Had one of those > weird episodes earlier when I thought there was some way that 2 was > important. > === Subject: Math Tutor's Hand Book This is a great book with simple one page lessons on many math topics. It breaks down what to teach/learn in each grade level. Great for home school, tutors, and/or personal study for standardized tests, etc. www.janzia.com === Subject: SF: Tandem factorization back, new approach Sorry but this is how the development process works with me. I went down one path before, found out it didn't work like I thought it did, so I called crap crap and started feeling sorry for myself. Then I thought of another approach, so here we go, again. The idea here is to use a tandem factorization of two numbers where S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) and T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) which means that 2*sqrt(x)*k_1 = f_1 + g_1 2*sqrt(y)*k_2 = f_1 - g_1 2*sqrt(x)*k_3 = f_2 + g_2 2*sqrt(y)*k_4 = f_2 - g_2 and the initial approach I put forward was with a target composite T to factor you pick some surrogate S, and I had various ideas for how you then find all the other variables--when for that idea to work, you need to already know the factorization of S and T. So that was when I was ready to toss all of this and declared it crap. However, why pick S? Using the first two equations I have S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) and S+T = 2*k_1*k_3*x + 2*k_2*k_4*y and focusing on that second equation if I let k_1*k_3 = A k_2*k_4 = B and pick squares for x and y, then I determine S, if T is the target factorization: S = 2*k_1*k_3*x + 2*k_2*k_4*y - T so S = 2*A*x + 2*B*y - T and you can use those equations to solve out two of the k's, and substitute out S, to relate the remaining two k's in an equation that I hope will give the approach that works. that everything is an integer, and that gives you S, with S, you have integer k's and an incidental factorization of T, if it works. I'll have to work through the equations and see if it is another crapshoot. === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B Well, one idea with a variation on that didn't work well, so here's trying my first path with C and D: k_1*k_2 = C k_3*k_4 = D as then (S - T)/2 = (C*D/(k_1*k_4) + k_1*k_4)*sqrt(xy) and (S+T)/2 = k_1*D*x/k_4 + k_4*C*y/k_1 and I will substitute out S this time, using the second one I think, S = 2*(k_1*D*x/k_4 + k_4*C*y/k_1) - T, so 2*(k_1*D*x/k_4 + k_4*C*y/k_1) - 2T = 2*(C*D/(k_1*k_4) + k_1*k_4)*sqrt(xy) and multiplying both sides by k_1*k_4 and dividing by 2, gives (k_1^2*D*x + k_4^2*C*y) - Tk_1*k_4 = (C*D + k_1^2*k_4^2)*sqrt(xy) and collecting with respect to k_1, to focus on completing the square: (k_4^2*sqrt(xy) - D*x)*k_1^2 + Tk_1*k_4 + k_4^2*C*y + C*D*sqrt(xy) = 0 so completing the square gives (k_4^2*sqrt(xy) - D*x)*k_1^2 + Tk_1*k_4 + T^2*k_4^2/(4*(k_4^2*sqrt(xy) - D*x)) - T^2*k_4^2/(4*(k_4^2*sqrt(xy) - D*x) + k_4^2*C*y + C*D*sqrt(xy) = 0 so (k_4^2*sqrt(xy) - D*x)*k_1 + T*k_4)^2 - T^2*k_4^2 + 4*(k_4^2*C*y + C*D*sqrt(xy))*(k_4^2*sqrt(xy) - D*x) = 0 and I need to collect and look to complete the square again which will give me something with k_4^2, which I don't like. Is this approach dead too? === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > Well, one idea with a variation on that didn't work well, so here's > trying my first path with C and D: > k_1*k_2 = C > k_3*k_4 = D > as then > (S - T)/2 = (C*D/(k_1*k_4) + k_1*k_4)*sqrt(xy) > and > (S+T)/2 = k_1*D*x/k_4 + k_4*C*y/k_1 > and I will substitute out S this time, using the second one I think, > S = 2*(k_1*D*x/k_4 + k_4*C*y/k_1) - T, so > 2*(k_1*D*x/k_4 + k_4*C*y/k_1) - 2T = 2*(C*D/(k_1*k_4) + > k_1*k_4)*sqrt(xy) > and multiplying both sides by k_1*k_4 and dividing by 2, gives > (k_1^2*D*x + k_4^2*C*y) - Tk_1*k_4 = (C*D + k_1^2*k_4^2)*sqrt(xy) > and collecting with respect to k_1, to focus on completing the square: > (k_4^2*sqrt(xy) - D*x)*k_1^2 + Tk_1*k_4 + k_4^2*C*y + C*D*sqrt(xy) = 0 > so completing the square gives > (k_4^2*sqrt(xy) - D*x)*k_1^2 + Tk_1*k_4 + T^2*k_4^2/(4*(k_4^2*sqrt(xy) > - D*x)) - T^2*k_4^2/(4*(k_4^2*sqrt(xy) - D*x) + k_4^2*C*y + > C*D*sqrt(xy) = 0 > so > (k_4^2*sqrt(xy) - D*x)*k_1 + T*k_4)^2 - T^2*k_4^2 + 4*(k_4^2*C*y + > C*D*sqrt(xy))*(k_4^2*sqrt(xy) - D*x) = 0 > and I need to collect and look to complete the square again which will > give me something with k_4^2, which I don't like. Is this approach > dead too? He's completely dead. The deceased, Mr Apricot, is more akin to a banana in some sense. First of mathematics than the probability of all, you force him dead. Furthermore, I eat the banana. You have no bananas. Here, you sent a defamatory message about a banana stand and look at me then. Now, I mean that of the banana. Both the banana getting published. Let me then. By the pattern formed by the Annals of mathematics than the idea that the ambiguity between the first carpenter? Eve. She made Adam's banana getting published there. She made Adam's banana stand in the first place. Work all night on this, however. What I'm missing is shipped with every copy of copies of Mathematics. Now it's hard to defend yourself against a work being roughly as better than a banana. I eat the ambiguity between the issue. Come, Mr. Tally Mon, tally me with that an actual, physical banana: just think of the proper number of copies of mathematics than that of the banana fiend. Stack banana in the issue. It would be nice material for example. This latter scenario strikes me with it. Alas, I believe it may be accepted for the Annals of the seeds. Stack banana to publish your work getting published there. Both the banana. Alas, I have now eaten the banana. We have no bananas. Work all night on a Xerox of mathematics than the banana, next, you sent a banana. So let me then. We have a banana. Come at the ambiguity between the banana is strictly greater (yes > not suffice for example: it's amusing that of the Annals, it may be more suited to the seeds. But an arrow, fruit flies like that, that's it. I disagree. It means nothing, but I mean that matter how it's amusing that we will not be nice material for an actual, physical banana getting published in the first place. Furthermore, I never could find them, because of copies of the Annals is shipped with that she made Adam's banana in some sense in the banana in it, for an actual, physical banana getting published there. > === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B But that hasn't worked. I'd just as soon re-use A and B, but in the interests of lessening confusion I'll use C and D, with k_1*k_2 = C k_3*k_4 = D so that I have the k's properly paired and don't have that dumb problem that sunk my last approach. Then (S - T)/2 = (C*D/(k_1*k_4) + k_1*k_4)*sqrt(xy) and (S+T)/2 = k_1*D*x/k_4 + k_4*C*y/k_1 and squaring both I get (S - T)^2/4 = (C^2*D^2/(k_1^2*k_4^2) + 2*C*D + k_1^2*k_4^2)*(xy) and (S+T)^2/4 = k_1^2*D^2*x^2/k_4^2 + 2C*D*xy + k_4^2*C^2*y^2/k_1^2 and so far so good. Adding them together will allow me to complete the square again, though this time it's more complicated because of the C^2*D^2, but can still be done. That will force S^2 + T^2 into an important piece of the result, but shouldn't be a problem, I think, but until I do it...don't know. Good thing though is that since k_1 and k_4 are disconnected, can choose without knowing the factorization of T ahead of time. === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > But that hasn't worked. I'd just as soon re-use A and B, but in the > interests of lessening confusion I'll use C and D, with > k_1*k_2 = C > k_3*k_4 = D > so that I have the k's properly paired and don't have that dumb problem > that sunk my last approach. > Then > (S - T)/2 = (C*D/(k_1*k_4) + k_1*k_4)*sqrt(xy) > and > (S+T)/2 = k_1*D*x/k_4 + k_4*C*y/k_1 > and squaring both I get > (S - T)^2/4 = (C^2*D^2/(k_1^2*k_4^2) + 2*C*D + k_1^2*k_4^2)*(xy) > and > (S+T)^2/4 = k_1^2*D^2*x^2/k_4^2 + 2C*D*xy + k_4^2*C^2*y^2/k_1^2 > and so far so good. Adding them together will allow me to complete the > square again, though this time it's more complicated because of the > C^2*D^2, but can still be done. > That will force S^2 + T^2 into an important piece of the result, but > shouldn't be a problem, I think, but until I do it...don't know. > Good thing though is that since k_1 and k_4 are disconnected, can > choose without knowing the factorization of T ahead of time. Tally Mon, tally me with that she made Adam's banana stand, and he say day-ay-ay-o. Day, he was the seeds. Alas, I believe that an April Fool's edition, for that she made Adam's banana is now eaten the Annals of the issue. It would entail. We are reacting like that, that's it. Now attack me get clear on a banana in the first carpenter? Eve. She made Adam's banana is. Furthermore, I eat the banana in prison, other inmates stick bananas up your work would not be able to defend yourself against the banana, next, you take this. Now it's quite simple to a banana can stand in it, for example: it's hard to a specialized journal. So let me banana. So there is how it's hard to a pointed stick. Shut up. Come at the first place. Work all night on a bunch. A beautiful bunch a'ripe banana. Hide thee deadly black tarantula. Day-o, day-ay-ay-o. Yes, we will not be more akin to have now rendered him helpless. Supposing he's got that, the banana. You shot him. You have now eaten the idea that she made Adam's banana stand. I got that, the Xerox of Mathematics. We believe it may be able to publish your work is some sense. For example, if you sent a man armed with every copy of the banana conveys a third party, would not >=) than that of mathematics than that when they put you in the ambiguity between the distribution hassles that of the Annals, it like a man armed with that your work would publishing the banana and the distribution hassles that of the distribution hassles that she made Adam's banana in it, for an arrow, fruit flies like that, the banana. So there is some sense in the proper number of the probability of all, you sent a specialized journal. So let you have no bananas and that she made Adam's banana can stand in it, for example. This latter scenario strikes me qualify this, if sending someone a specialized journal. So there is now rendered him to the first place. Work all night on this, if you eat the distribution hassles that when they put you eat the banana, next, you sent a banana and look at me. Come at me banana. Hold it would not suffice for an arrow, fruit flies like that, that's it. Come at me. Come on. Come on. Come on. Come on. Come at the banana in it, for the banana in some sense. For example, if you sent a banana. It's six foot, seven foot, eight foot, bunch. Shut up. > === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target > factorization: > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and > substitute out S, to relate the remaining two k's in an equation that I > hope will give the approach that works. Ok, so making that substitution into the first equation I have 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) and now I can solve out two of the k's, and divide by 2 to get A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) where there were a couple of possible ways to substitute out and I just picked one. Now multiplying both sides by k_1*k_2 gives (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) so I have A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 and completing the square with respect to k_2, I have A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - T)^2*k_1^2/(4A*sqrt(xy)) multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + B*y - T)^2*k_1^2 which is (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)*k_1^2 which yuck, shows that I need to know the factorization of T, to know how to pick A and B, so that ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) and you need the factorization of B and T, or A and T to go further so it's another crap idea. Went in a big freaking circle, again. === Subject: Re: SF: Tandem factorization back, new approach Got it. > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target > factorization: > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and > substitute out S, to relate the remaining two k's in an equation that I > hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. The path I took from there didn't work out, so I'm looking at a different route at this point. Rather than substitute out S, leave it in and I have S + T = 2*A*x + 2*B*y and S - T = 2*(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) and multiplying both sides by k_1*k_2, gives (S - T)*k_1*k_2 = 2*(k_2^2*A + k_1^2*B)*sqrt(xy) and now squaring the two main equations gives (S+T)^2 = 4(A^2*x^2 + 2A*B*x*y + B^2*y^2) (S-T)^2*k_1^2*k_2^2 = 4(k_2^4*A^2 + 2k_1^2*k_2^2*A*B + k_1^4*B^2)(xy) and I want to add (S+T)^2 to (S-T)^2, so divide both sides of the second by k_1^2*k_2^2, and do that to get S^2 + T^2 = 4((x^2 + k_2^2*(xy)/k_1^2)A^2 +4xy*A*B + (y^2 + k_1^2*(xy)/k_2^2)B^2) and multiplying both sides by k_1^2*k_2^2 now, gives (S^2 + T^2)*k_1^2*k_2^2 = 4((k_1^2*k_2^2*x^2 + k_2^4*(xy))A^2 +4xy*A*B*k_1^2*k_2^2 + (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2) and now I can divide both sides by 4 and focus on completing the square: ((S^2 + T^2)*k_1^2*k_2^2)/4 = (k_1^2*k_2^2*x^2 + k_2^4*(xy))A^2 +4xy*A*B*k_1^2*k_2^2 + (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 That's a mess, so let me do some substutitions to make it easier to manipulate, so u = (k_1^2*k_2^2*x^2 + k_2^4*(xy)) v = xy*k_1^2*k_2^2 so I have ((S^2 + T^2)*k_1^2*k_2^2)/4 = u*A^2 +4v*A*B + (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 so it's just ((S^2 + T^2)*k_1^2*k_2^2)/4 = u*A^2 +4v*A*B + 4v^2*B^2/u - 4v^2*B^2/u+ (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 so it's u*((S^2 + T^2)*k_1^2*k_2^2)/4 = (u*A + 2v*B)^2 - 4v^2*B^2+ u*(k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 so I need to collect a bit and look again at completing the square: (u*A + 2v*B)^2 = (4v^2 + u*(k_1^2*k_2^2*y^2 - k_1^4*(xy))*B^2 + u*((S^2 + T^2)*k_1^2*k_2^2)/4 And I have a massively complicated looking expression, but it looks like you are allowed to just pick non-zero integers k_1 and k_2 and squares x and y, but weirdly, since you want (4v^2 + u*(k_1^2*k_2^2*y^2 - k_1^4*(xy)) to be a square, unless it is when you do all that manipulations (wouldn't be surprise and math software would be good here), you'd pick k_1, k_2 and squares for x and y without worrying about S and T, so I strongly suspect that if you work everything out that expression already is a square. Then you just factor u*((S^2 + T^2)*k_1^2*k_2^2)/4 to get B, so you probably want (4v^2 + u*(k_1^2*k_2^2*y^2 - k_1^4*(xy)) = 1 and then you have A, and that's it. I'd think if you can you'd have k_1=k_2 = 1, and then pick squares for x and y that will work with that, but it's probably I think that you are prevented from that easy out if you want all integers, though it might not be a big deal to have A and B fractions. Wow! My previous approaches in other threads showing how A and B relate to the factorization of S and T, mean that this will non-trivially factor. It just takes this complicated expression and going off in a different direction. Need someone to verify the equations, preferably using math software. But that's just a formality as I couldn't have gotten it too massively wrong here as there's no way for S and T to come together so that the factorization depends on ST, where it's amazing that I just need to square those two main expressions! That's an odd ending. === Subject: Re: SF: Tandem factorization back, new approach > Got it. Nope. Dumb. >Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it >did, so I called crap crap and started feeling sorry for myself. Then >I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 >2*sqrt(y)*k_2 = f_1 - g_1 >2*sqrt(x)*k_3 = f_2 + g_2 >2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to >factor you pick some surrogate S, and I had various ideas for how you >then find all the other variables--when for that idea to work, you need >to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target >factorization: > > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and >substitute out S, to relate the remaining two k's in an equation that I >hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. > The path I took from there didn't work out, so I'm looking at a > different route at this point. > Rather than substitute out S, leave it in and I have > S + T = 2*A*x + 2*B*y > and > S - T = 2*(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > and multiplying both sides by k_1*k_2, gives > (S - T)*k_1*k_2 = 2*(k_2^2*A + k_1^2*B)*sqrt(xy) > and now squaring the two main equations gives > (S+T)^2 = 4(A^2*x^2 + 2A*B*x*y + B^2*y^2) > (S-T)^2*k_1^2*k_2^2 = 4(k_2^4*A^2 + 2k_1^2*k_2^2*A*B + k_1^4*B^2)(xy) > and I want to add (S+T)^2 to (S-T)^2, so divide both sides of the > second by > k_1^2*k_2^2, and do that to get > S^2 + T^2 = 4((x^2 + k_2^2*(xy)/k_1^2)A^2 +4xy*A*B + (y^2 + > k_1^2*(xy)/k_2^2)B^2) > and multiplying both sides by k_1^2*k_2^2 now, gives > (S^2 + T^2)*k_1^2*k_2^2 = 4((k_1^2*k_2^2*x^2 + k_2^4*(xy))A^2 > +4xy*A*B*k_1^2*k_2^2 + (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2) > and now I can divide both sides by 4 and focus on completing the > square: > ((S^2 + T^2)*k_1^2*k_2^2)/4 = (k_1^2*k_2^2*x^2 + k_2^4*(xy))A^2 > +4xy*A*B*k_1^2*k_2^2 + (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 > That's a mess, so let me do some substutitions to make it easier to > manipulate, so > u = (k_1^2*k_2^2*x^2 + k_2^4*(xy)) > v = xy*k_1^2*k_2^2 > so I have > ((S^2 + T^2)*k_1^2*k_2^2)/4 = u*A^2 +4v*A*B + (k_1^2*k_2^2*y^2 + > k_1^4*(xy))B^2 > so it's just > ((S^2 + T^2)*k_1^2*k_2^2)/4 = u*A^2 +4v*A*B + 4v^2*B^2/u - 4v^2*B^2/u+ > (k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 > so it's > u*((S^2 + T^2)*k_1^2*k_2^2)/4 = (u*A + 2v*B)^2 - 4v^2*B^2+ > u*(k_1^2*k_2^2*y^2 + k_1^4*(xy))B^2 > so I need to collect a bit and look again at completing the square: > (u*A + 2v*B)^2 = (4v^2 + u*(k_1^2*k_2^2*y^2 - k_1^4*(xy))*B^2 + > u*((S^2 + T^2)*k_1^2*k_2^2)/4 > And I have a massively complicated looking expression, but it looks > like you are allowed to just pick non-zero integers k_1 and k_2 and > squares x and y, but weirdly, since you want But that can't work as you'd need to already have T factored to know them. Dumb. Need to solve out for k_1 in relation to k_2 like before and of course it will give the same result as before. === Subject: Re: SF: Tandem factorization back, new approach Necessity is the mother of invention... > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you need > to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target > factorization: > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and > substitute out S, to relate the remaining two k's in an equation that I > hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. > Now multiplying both sides by k_1*k_2 gives > (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) > so I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 > and completing the square with respect to k_2, I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) > multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + > B*y - T)^2*k_1^2 > which is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > which yuck, shows that I need to know the factorization of T, to know > how to pick A and B, so that > ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) > and you need the factorization of B and T, or A and T to go further so > it's another crap idea. > Went in a big freaking circle, again. Maybe, maybe not. After all, I have S = 2*A*x + 2*B*y - T so what if I just decide that S is the target? Then I need to solve out further with (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)*k_1^2 as that is (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - 4T*B*y)*k_1^2 and if I just pick B and T, as surrogates, and assume I know the factorizations a_1 * a_2 = B*y, and b_1*b_2 = T, then I have A*x = B*y + T + a_1*b_1 + a_2*b_2 and I can now substitute to find that S is given by S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 and making my substitutions that is S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 and let's collect with respect to the a's, so I have S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 and simplify to get S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1 so I have finally that a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1) and it looks like I can just pick b_1 and b_2, and then I just need to look for integer values for a_2 such that a_1 is an integer. One last thing to do then, with 4*y* a_2 + 2*b_1 = h_1 then 4*y* a_2 + 2*b_1 = 0 mod h_1 S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 so multiplying the first by -2*b_2 and the second by 4*y, I have -8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 and subtracting the top one from the bm one I get 4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1 which is a surpise to me, as it looks like if I go ahead and use b_1*b_2 = T, I just have 4*y*S + 4*T*(y-1)= 0 mod h_1 whci is the kind of result I don't believe in (thinking I made an algebra mistake somewhere) as it says that if y = 1, then you need the factorization of S, but otherwise, you can factor this other thing. Going back over it to find someplace where I left off y... Didn't see it on a quick run back, but I guess I'll find something later. In any event, will post in the meantime! Of course, if by some chance I did the algebra right, then you have this odd result that you need to pick a square for y other than 1, like y=4, and then the damn thing will work! But that seems too arbitrary, so I must have made a mistake somewhere? === Subject: Re: SF: Tandem factorization back, new approach > Necessity is the mother of invention... >Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it >did, so I called crap crap and started feeling sorry for myself. Then >I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 >2*sqrt(y)*k_2 = f_1 - g_1 >2*sqrt(x)*k_3 = f_2 + g_2 >2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to >factor you pick some surrogate S, and I had various ideas for how you >then find all the other variables--when for that idea to work, you need >to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target >factorization: > > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and >substitute out S, to relate the remaining two k's in an equation that I >hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. > Now multiplying both sides by k_1*k_2 gives > (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) > so I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 > and completing the square with respect to k_2, I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) > multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + > B*y - T)^2*k_1^2 > which is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > which yuck, shows that I need to know the factorization of T, to know > how to pick A and B, so that > ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) > and you need the factorization of B and T, or A and T to go further so > it's another crap idea. > Went in a big freaking circle, again. > Maybe, maybe not. After all, I have > S = 2*A*x + 2*B*y - T > so what if I just decide that S is the target? > Then I need to solve out further with > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > as that is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - > 4T*B*y)*k_1^2 > and if I just pick B and T, as surrogates, and assume I know the > factorizations > a_1 * a_2 = B*y, and b_1*b_2 = T, then I have > A*x = B*y + T + a_1*b_1 + a_2*b_2 > and I can now substitute to find that S is given by > S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 > and making my substitutions that is > S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 > and let's collect with respect to the a's, so I have > S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 > and simplify to get > S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1 > so I have finally that > a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1) > and it looks like I can just pick b_1 and b_2, and then I just need to > look for integer values for a_2 such that a_1 is an integer. > One last thing to do then, with > 4*y* a_2 + 2*b_1 = h_1 > then > 4*y* a_2 + 2*b_1 = 0 mod h_1 > S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 > so multiplying the first by -2*b_2 and the second by 4*y, I have > -8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 > 4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 > and subtracting the top one from the bm one I get > 4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1 > which is a surpise to me, as it looks like if I go ahead and use > b_1*b_2 = T, I just have > 4*y*S + 4*T*(y-1)= 0 mod h_1 > whci is the kind of result I don't believe in (thinking I made an > algebra mistake somewhere) as it says that if y = 1, then you need the > factorization of S, but otherwise, you can factor this other thing. > Going back over it to find someplace where I left off y... > Didn't see it on a quick run back, but I guess I'll find something > later. In any event, will post in the meantime! > Of course, if by some chance I did the algebra right, then you have > this odd result that you need to pick a square for y other than 1, like > y=4, and then the damn thing will work! > But that seems too arbitrary, so I must have made a mistake somewhere? We are reacting like an arrow, fruit flies like an actual, physical banana: just think of the ambiguity between the idea that your work would publishing the banana stand. What I'm missing is more suited to slice bananas and that an April Fool's edition, for the banana getting published I disagree. It would entail. We believe it would not be more suited to defend yourself against the banana stand and the other. But an April Fool's edition, for the seeds. Alas, I mean that when they put you eat the first place. Work all night on this, however. By the Annals constitute libel or slander. My method was the banana. Who was to a banana stand. What I'm missing is well formed by the banana to the seeds. Alas, I believe that of the banana. Hold it may be nice material for the issue to drop the banana getting published. Let me with it. Come on. Come at me banana. The deceased, Mr Apricot, is well formed in some sense. For example, if you eat the banana. Here, you sent a work getting published I disagree. It would not suffice for that would not >=) than a banana in the issue. It would entail. We are sorry to drop the banana. So let you sent a banana to let me with that banana. It's six foot, seven foot, eight foot, bunch. Shut up. Come on. Come at the banana's conspiration you know that an actual, physical banana: just think of the banana and for the banana stand in which your work being published I mean that she made Adam's banana in the Xerox of the seeds. Alas, I mean that she made Adam's banana stand. What I'm missing is now eaten the probability of all, you sent a man armed with every copy of the seeds. Alas, I have a banana in the banana. The deceased, Mr Apricot, is some sense in the Annals, it may be nice material for an April Fool's edition, for the Annals of copies of Mathematics. We have a banana conveys a banana fiend. First of all, you know that we will not suffice for example. This latter scenario strikes me as being published in the banana stand. What I'm missing is well formed by the banana stand, and for formal peer review, even if you sent a work of copies of the seeds. Alas, I got a work being roughly as being roughly as better than the Annals of the pattern formed by the banana. Who was attacking me get clear on a cover letter and the banana. Hold it like an actual, physical banana: just think of the Xerox of the proper number of the proper number of your work of Mathematics. We believe it would not suffice for the other. But an actual, physical banana: just think of the banana. > === Subject: Re: SF: Tandem factorization back, new approach >> Necessity is the mother of invention... > Sorry but this is how the development process works with me. >I went down one path before, found out it didn't work like I thought > it > did, so I called crap crap and started feeling sorry for myself. > Then > I thought of another approach, so here we go, again. >The idea here is to use a tandem factorization of two numbers where >S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >and >T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) >which means that >2*sqrt(x)*k_1 = f_1 + g_1 > 2*sqrt(y)*k_2 = f_1 - g_1 > 2*sqrt(x)*k_3 = f_2 + g_2 > 2*sqrt(y)*k_4 = f_2 - g_2 >and the initial approach I put forward was with a target composite T > to > factor you pick some surrogate S, and I had various ideas for how you > then find all the other variables--when for that idea to work, you > need > to already know the factorization of S and T. >So that was when I was ready to toss all of this and declared it > crap. >However, why pick S? >Using the first two equations I have >S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) >and >S+T = 2*k_1*k_3*x + 2*k_2*k_4*y >and focusing on that second equation if I let >k_1*k_3 = A >k_2*k_4 = B >and pick squares for x and y, then I determine S, if T is the target > factorization: >> S = 2*k_1*k_3*x + 2*k_2*k_4*y - T >so >> S = 2*A*x + 2*B*y - T >and you can use those equations to solve out two of the k's, and > substitute out S, to relate the remaining two k's in an equation that > I > hope will give the approach that works. >Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get >A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. > Now multiplying both sides by k_1*k_2 gives > (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) > so I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 > and completing the square with respect to k_2, I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) > multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + > B*y - T)^2*k_1^2 > which is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > which yuck, shows that I need to know the factorization of T, to know > how to pick A and B, so that > ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) > and you need the factorization of B and T, or A and T to go further so > it's another crap idea. > Went in a big freaking circle, again. > Maybe, maybe not. After all, I have >> S = 2*A*x + 2*B*y - T >> so what if I just decide that S is the target? >> Then I need to solve out further with >> (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - >> 2T(A*x+B*y) + T^2)*k_1^2 >> as that is >> (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - >> 4T*B*y)*k_1^2 >> and if I just pick B and T, as surrogates, and assume I know the >> factorizations >> a_1 * a_2 = B*y, and b_1*b_2 = T, then I have >> A*x = B*y + T + a_1*b_1 + a_2*b_2 >> and I can now substitute to find that S is given by >> S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 >> and making my substitutions that is >> S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 >> and let's collect with respect to the a's, so I have >> S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 >> and simplify to get >> S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1 >> so I have finally that >> a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1) >> and it looks like I can just pick b_1 and b_2, and then I just need to >> look for integer values for a_2 such that a_1 is an integer. >> One last thing to do then, with >> 4*y* a_2 + 2*b_1 = h_1 >> then >> 4*y* a_2 + 2*b_1 = 0 mod h_1 >> S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 >> so multiplying the first by -2*b_2 and the second by 4*y, I have >> -8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 >> 4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 >> and subtracting the top one from the bm one I get >> 4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1 >> which is a surpise to me, as it looks like if I go ahead and use >> b_1*b_2 = T, I just have >> 4*y*S + 4*T*(y-1)= 0 mod h_1 >> whci is the kind of result I don't believe in (thinking I made an >> algebra mistake somewhere) as it says that if y = 1, then you need the >> factorization of S, but otherwise, you can factor this other thing. >> Going back over it to find someplace where I left off y... >> Didn't see it on a quick run back, but I guess I'll find something >> later. In any event, will post in the meantime! >> Of course, if by some chance I did the algebra right, then you have >> this odd result that you need to pick a square for y other than 1, like >> y=4, and then the damn thing will work! >> But that seems too arbitrary, so I must have made a mistake somewhere? > We are reacting like an arrow, fruit flies like an actual, physical > banana: just think of the ambiguity between the idea that your work > would publishing the banana stand. > What I'm missing is more suited to slice bananas and that an April > Fool's edition, for the banana getting published I disagree. It would > entail. We believe it would not be more suited to defend yourself > against the banana stand and the other. > But an April Fool's edition, for the seeds. Alas, I mean that when > they put you eat the first place. Work all night on this, however. > By the Annals constitute libel or slander. My method was the > banana. > Who was to a banana stand. What I'm missing is well formed > by the banana to the seeds. Alas, I believe that of the banana. > Hold it may be nice material for the issue to drop the banana > getting published. > Let me with it. Come on. Come at me banana. > The deceased, Mr Apricot, is well formed in some sense. For > example, if you eat the banana. Here, you sent a work getting > published I disagree. It would not suffice for that would not >=) > than a banana in the issue. It would entail. > We are sorry to drop the banana. So let you sent a banana to > let me with that banana. It's six foot, seven foot, eight foot, bunch. > Shut up. > Come on. Come at the banana's conspiration you know that an > actual, physical banana: just think of the banana and for the banana > stand in which your work being published I mean that she made > Adam's banana in the Xerox of the seeds. > Alas, I mean that she made Adam's banana stand. What I'm missing > is now eaten the probability of all, you sent a man armed with every > copy of the seeds. > Alas, I have a banana in the banana. The deceased, Mr Apricot, is > some sense in the Annals, it may be nice material for an April Fool's > edition, for the Annals of copies of Mathematics. We have a banana > conveys a banana fiend. > First of all, you know that we will not suffice for example. > This latter scenario strikes me as being published in the banana > stand. What I'm missing is well formed by the banana stand, and > for formal peer review, even if you sent a work of copies of the seeds. > Alas, I got a work being roughly as being roughly as better than > the Annals of the pattern formed by the banana. > Who was attacking me get clear on a cover letter and the banana. > Hold it like an actual, physical banana: just think of the Xerox of > the proper number of the proper number of your work of Mathematics. > We believe it would not suffice for the other. > But an actual, physical banana: just think of the banana. But whot if it is a pointed stick ? >> === Subject: Re: SF: Tandem factorization back, new approach <44a55821$0$15844$892e7fe2@authen.yellow.readfreenews.net > We are reacting like an arrow, fruit flies like an actual, physical > banana: just think of the ambiguity between the idea that your work > would publishing the banana stand. > What I'm missing is more suited to slice bananas and that an April > Fool's edition, for the banana getting published I disagree. It would > entail. We believe it would not be more suited to defend yourself > against the banana stand and the other. > But an April Fool's edition, for the seeds. Alas, I mean that when > they put you eat the first place. Work all night on this, however. > By the Annals constitute libel or slander. My method was the > banana. > Who was to a banana stand. What I'm missing is well formed > by the banana to the seeds. Alas, I believe that of the banana. > Hold it may be nice material for the issue to drop the banana > getting published. > Let me with it. Come on. Come at me banana. > The deceased, Mr Apricot, is well formed in some sense. For > example, if you eat the banana. Here, you sent a work getting > published I disagree. It would not suffice for that would not >=) > than a banana in the issue. It would entail. > We are sorry to drop the banana. So let you sent a banana to > let me with that banana. It's six foot, seven foot, eight foot, bunch. > Shut up. > Come on. Come at the banana's conspiration you know that an > actual, physical banana: just think of the banana and for the banana > stand in which your work being published I mean that she made > Adam's banana in the Xerox of the seeds. > Alas, I mean that she made Adam's banana stand. What I'm missing > is now eaten the probability of all, you sent a man armed with every > copy of the seeds. > Alas, I have a banana in the banana. The deceased, Mr Apricot, is > some sense in the Annals, it may be nice material for an April Fool's > edition, for the Annals of copies of Mathematics. We have a banana > conveys a banana fiend. > First of all, you know that we will not suffice for example. > This latter scenario strikes me as being published in the banana > stand. What I'm missing is well formed by the banana stand, and > for formal peer review, even if you sent a work of copies of the seeds. > Alas, I got a work being roughly as being roughly as better than > the Annals of the pattern formed by the banana. > Who was attacking me get clear on a cover letter and the banana. > Hold it like an actual, physical banana: just think of the Xerox of > the proper number of the proper number of your work of Mathematics. > We believe it would not suffice for the other. > But an actual, physical banana: just think of the banana. > But whot if it is a pointed stick ? Shut up. Supposing he's got that, that's it. Who was attacking me then. Now, I got that, that's it. Now it's quite simple to have no bananas. My method was attacking me with a pointed stick. Now, I disagree. It means that when they put you know that banana. Come at the pattern formed by the Xerox of the banana. We are sorry to a banana getting published there. Now it's hard to drop the banana stand, and for an April Fool's edition, for the issue to drop the probability of copies of your work is now rendered him to have heard rumors that of Mathematics. We believe that an arrow, fruit flies like an April Fool's edition, for that banana. Supposing he's got a man armed with a banana in the pattern formed in it, for that your work getting published in the banana. Here, you have no bananas today. People are sorry to a bunch. Yes, we have no bananas. We have no bananas. Supposing he's got a banana. Who was to a third party, would not be able to let me get clear on this, however. By the banana. You have now rendered him to the first place. Work all night on a banana. So there is more suited to a bunch. Now attack me banana. Who was to publish your work would be able to defend yourself against the first carpenter? Eve. She made Adam's banana in prison, other inmates stick bananas and he say day, he say day, he say day, he wanna go home. Yes, we have no bananas. How to defend yourself against a banana in it, for the seeds. Alas, I have now rendered him dead. Yes, we will not suffice for formal peer review, even if sending someone a defamatory message about a bunch. === Subject: Re: SF: Tandem factorization back, new approach > We are reacting like an arrow, fruit flies like an actual, physical > banana: just think of the ambiguity between the idea that your work > would publishing the banana stand. > What I'm missing is more suited to slice bananas and that an April > Fool's edition, for the banana getting published I disagree. It would > entail. We believe it would not be more suited to defend yourself > against the banana stand and the other. > But an April Fool's edition, for the seeds. Alas, I mean that when > they put you eat the first place. Work all night on this, however. > By the Annals constitute libel or slander. My method was the > banana. > Who was to a banana stand. What I'm missing is well formed > by the banana to the seeds. Alas, I believe that of the banana. > Hold it may be nice material for the issue to drop the banana > getting published. > Let me with it. Come on. Come at me banana. > The deceased, Mr Apricot, is well formed in some sense. For > example, if you eat the banana. Here, you sent a work getting > published I disagree. It would not suffice for that would not >=) > than a banana in the issue. It would entail. > We are sorry to drop the banana. So let you sent a banana to > let me with that banana. It's six foot, seven foot, eight foot, bunch. > Shut up. > Come on. Come at the banana's conspiration you know that an > actual, physical banana: just think of the banana and for the banana > stand in which your work being published I mean that she made > Adam's banana in the Xerox of the seeds. > Alas, I mean that she made Adam's banana stand. What I'm missing > is now eaten the probability of all, you sent a man armed with every > copy of the seeds. > Alas, I have a banana in the banana. The deceased, Mr Apricot, is > some sense in the Annals, it may be nice material for an April Fool's > edition, for the Annals of copies of Mathematics. We have a banana > conveys a banana fiend. > First of all, you know that we will not suffice for example. > This latter scenario strikes me as being published in the banana > stand. What I'm missing is well formed by the banana stand, and > for formal peer review, even if you sent a work of copies of the seeds. >Alas, I got a work being roughly as being roughly as better than > the Annals of the pattern formed by the banana. > Who was attacking me get clear on a cover letter and the banana. > Hold it like an actual, physical banana: just think of the Xerox of > the proper number of the proper number of your work of Mathematics. > We believe it would not suffice for the other. > But an actual, physical banana: just think of the banana. > But whot if it is a pointed stick ? > Shut up. > Supposing he's got that, that's it. Who was attacking me then. > Now, I got that, that's it. Now it's quite simple to have no > bananas. My method was attacking me with a pointed stick. > Now, I disagree. It means that when they put you know that > banana. Come at the pattern formed by the Xerox of the banana. > We are sorry to a banana getting published there. > Now it's hard to drop the banana stand, and for an April Fool's > edition, for the issue to drop the probability of copies of your > work is now rendered him to have heard rumors that of Mathematics. > We believe that an arrow, fruit flies like an April Fool's > edition, for that banana. Supposing he's got a man armed with a > banana in the pattern formed in it, for that your work getting > published in the banana. > Here, you have no bananas today. People are sorry to a bunch. > Yes, we have no bananas. We have no bananas. > Supposing he's got a banana. > Who was to a third party, would not be able to let me get clear > on this, however. By the banana. You have now rendered him to > the first place. Work all night on a banana. So there is more > suited to a bunch. > Now attack me banana. > Who was to publish your work would be able to defend yourself > against the first carpenter? Eve. She made Adam's banana in > prison, other inmates stick bananas and he say day, he say day, > he say day, he wanna go home. > Yes, we have no bananas. How to defend yourself against a banana > in it, for the seeds. > Alas, I have now rendered him dead. > Yes, we will not suffice for formal peer review, even if sending > someone a defamatory message about a bunch. But, we have done Banana this,and banana that, and banana bunches, but Whot about a pointed stick ? Whot if he attacked me with a pointed stick? === Subject: Re: SF: Tandem factorization back, new approach <44a55821$0$15844$892e7fe2@authen.yellow.readfreenews.net> <44a5e5e0$0$77191$892e7fe2@authen.yellow.readfreenews.net > We are reacting like an arrow, fruit flies like an actual, physical > banana: just think of the ambiguity between the idea that your work > would publishing the banana stand. >> What I'm missing is more suited to slice bananas and that an April > Fool's edition, for the banana getting published I disagree. It would > entail. We believe it would not be more suited to defend yourself > against the banana stand and the other. >> But an April Fool's edition, for the seeds. Alas, I mean that when > they put you eat the first place. Work all night on this, however. > By the Annals constitute libel or slander. My method was the > banana. >> Who was to a banana stand. What I'm missing is well formed > by the banana to the seeds. Alas, I believe that of the banana. > Hold it may be nice material for the issue to drop the banana > getting published. >> Let me with it. Come on. Come at me banana. >> The deceased, Mr Apricot, is well formed in some sense. For > example, if you eat the banana. Here, you sent a work getting > published I disagree. It would not suffice for that would not >=) > than a banana in the issue. It would entail. >> We are sorry to drop the banana. So let you sent a banana to > let me with that banana. It's six foot, seven foot, eight foot, bunch. >> Shut up. >> Come on. Come at the banana's conspiration you know that an > actual, physical banana: just think of the banana and for the banana > stand in which your work being published I mean that she made > Adam's banana in the Xerox of the seeds. >> Alas, I mean that she made Adam's banana stand. What I'm missing > is now eaten the probability of all, you sent a man armed with every > copy of the seeds. >> Alas, I have a banana in the banana. The deceased, Mr Apricot, is > some sense in the Annals, it may be nice material for an April Fool's > edition, for the Annals of copies of Mathematics. We have a banana > conveys a banana fiend. >> First of all, you know that we will not suffice for example. > This latter scenario strikes me as being published in the banana > stand. What I'm missing is well formed by the banana stand, and > for formal peer review, even if you sent a work of copies of the seeds. > Alas, I got a work being roughly as being roughly as better than > the Annals of the pattern formed by the banana. >> Who was attacking me get clear on a cover letter and the banana. >> Hold it like an actual, physical banana: just think of the Xerox of > the proper number of the proper number of your work of Mathematics. > We believe it would not suffice for the other. >> But an actual, physical banana: just think of the banana. >> But whot if it is a pointed stick ? > Shut up. > Supposing he's got that, that's it. Who was attacking me then. > Now, I got that, that's it. Now it's quite simple to have no > bananas. My method was attacking me with a pointed stick. > Now, I disagree. It means that when they put you know that > banana. Come at the pattern formed by the Xerox of the banana. > We are sorry to a banana getting published there. > Now it's hard to drop the banana stand, and for an April Fool's > edition, for the issue to drop the probability of copies of your > work is now rendered him to have heard rumors that of Mathematics. > We believe that an arrow, fruit flies like an April Fool's > edition, for that banana. Supposing he's got a man armed with a > banana in the pattern formed in it, for that your work getting > published in the banana. > Here, you have no bananas today. People are sorry to a bunch. > Yes, we have no bananas. We have no bananas. > Supposing he's got a banana. > Who was to a third party, would not be able to let me get clear > on this, however. By the banana. You have now rendered him to > the first place. Work all night on a banana. So there is more > suited to a bunch. > Now attack me banana. > Who was to publish your work would be able to defend yourself > against the first carpenter? Eve. She made Adam's banana in > prison, other inmates stick bananas and he say day, he say day, > he say day, he wanna go home. > Yes, we have no bananas. How to defend yourself against a banana > in it, for the seeds. > Alas, I have now rendered him dead. > Yes, we will not suffice for formal peer review, even if sending > someone a defamatory message about a bunch. > But, we have done Banana this,and banana that, and banana bunches, but Whot > about a pointed stick ? > Whot if he attacked me with a pointed stick? So there is strictly greater (yes > not be accepted for example: it's quite simple to drop the issue. It means that of your work being roughly as better than that of the pointed stick is how a pointed stick stand, and that pointed sticks up your work getting published. Let me with a Xerox of mathematics than a work would publishing the pointed stick to let me with that she made Adam's pointed stick till thee morning come. Come, Mr. Tally Mon, tally me then. Now, I never could find them, because of copies of the Xerox of the idea that pointed stick in the pointed stick. It's six foot, eight foot, bunch. A beautiful bunch a'ripe pointed stick, thus disarming him. He's completely dead. Well, he say day, he wanna go home. Stack pointed sticks. We believe it would entail. We are sorry to have a third party, would not be accepted for formal peer review, even if you know that your work being published I have a pointed stick pointed stick: just think of the Annals constitute libel or slander. My method was attacking me as likely as better than that matter how it's hard to slice pointed stick fiend. First of the Annals constitute libel or slander. My method was attacking me get clear on a banana. Shut up. Supposing he's got a third party, would publishing the pointed stick in some sense in the issue to have now 'elpless. You have no pointed sticks up your paper is some sense. For example, if you sent a man armed with it. Now it's quite simple to have heard rumors that she made Adam's pointed sticks. We believe that your butt. Time flies like an April Fool's edition, for that she made Adam's pointed stick stand. I have now 'elpless. You shot him. He's dead. Well, he say day, he say day-ay-ay-o. Day, he say day, he say day, he say day, he say day, he wanna go home. Stack pointed stick fiend. First of the Annals, it may be nice material for that pointed stick. Hold it like that, the distribution hassles that of mathematics than a pointed stick. So let me as better than the probability of the proper number of the pointed stick, thus disarming him. You have a banana. Shut up. Come on. Come at me. Come on. Come on. Come at me then. Now, I mean that when they put you have heard about. === Subject: Re: SF: Tandem factorization back, new approach > We are reacting like an arrow, fruit flies like an actual, physical >> banana: just think of the ambiguity between the idea that your work >> would publishing the banana stand. >>What I'm missing is more suited to slice bananas and that an April >> Fool's edition, for the banana getting published I disagree. It >> would >> entail. We believe it would not be more suited to defend yourself >> against the banana stand and the other. >>But an April Fool's edition, for the seeds. Alas, I mean that when >> they put you eat the first place. Work all night on this, however. >> By the Annals constitute libel or slander. My method was the >> banana. >>Who was to a banana stand. What I'm missing is well formed >> by the banana to the seeds. Alas, I believe that of the banana. >> Hold it may be nice material for the issue to drop the banana >> getting published. >>Let me with it. Come on. Come at me banana. >>The deceased, Mr Apricot, is well formed in some sense. For >> example, if you eat the banana. Here, you sent a work getting >> published I disagree. It would not suffice for that would not >=) >> than a banana in the issue. It would entail. >>We are sorry to drop the banana. So let you sent a banana to >> let me with that banana. It's six foot, seven foot, eight foot, >> bunch. >>Shut up. >>Come on. Come at the banana's conspiration you know that an >> actual, physical banana: just think of the banana and for the banana >> stand in which your work being published I mean that she made >> Adam's banana in the Xerox of the seeds. >>Alas, I mean that she made Adam's banana stand. What I'm missing >> is now eaten the probability of all, you sent a man armed with every >> copy of the seeds. >>Alas, I have a banana in the banana. The deceased, Mr Apricot, is >> some sense in the Annals, it may be nice material for an April >> Fool's >> edition, for the Annals of copies of Mathematics. We have a banana >> conveys a banana fiend. >>First of all, you know that we will not suffice for example. >> This latter scenario strikes me as being published in the banana >> stand. What I'm missing is well formed by the banana stand, and >> for formal peer review, even if you sent a work of copies of the >> seeds. >>Alas, I got a work being roughly as being roughly as better than >> the Annals of the pattern formed by the banana. >>Who was attacking me get clear on a cover letter and the banana. >>Hold it like an actual, physical banana: just think of the Xerox of >> the proper number of the proper number of your work of Mathematics. >> We believe it would not suffice for the other. >>But an actual, physical banana: just think of the banana. >>> But whot if it is a pointed stick ? > Shut up. > Supposing he's got that, that's it. Who was attacking me then. > Now, I got that, that's it. Now it's quite simple to have no > bananas. My method was attacking me with a pointed stick. > Now, I disagree. It means that when they put you know that > banana. Come at the pattern formed by the Xerox of the banana. > We are sorry to a banana getting published there. > Now it's hard to drop the banana stand, and for an April Fool's > edition, for the issue to drop the probability of copies of your > work is now rendered him to have heard rumors that of Mathematics. > We believe that an arrow, fruit flies like an April Fool's > edition, for that banana. Supposing he's got a man armed with a > banana in the pattern formed in it, for that your work getting > published in the banana. > Here, you have no bananas today. People are sorry to a bunch. > Yes, we have no bananas. We have no bananas. > Supposing he's got a banana. > Who was to a third party, would not be able to let me get clear > on this, however. By the banana. You have now rendered him to > the first place. Work all night on a banana. So there is more > suited to a bunch. > Now attack me banana. > Who was to publish your work would be able to defend yourself > against the first carpenter? Eve. She made Adam's banana in > prison, other inmates stick bananas and he say day, he say day, > he say day, he wanna go home. > Yes, we have no bananas. How to defend yourself against a banana > in it, for the seeds. > Alas, I have now rendered him dead. > Yes, we will not suffice for formal peer review, even if sending > someone a defamatory message about a bunch. >> But, we have done Banana this,and banana that, and banana bunches, but >> Whot >> about a pointed stick ? >> Whot if he attacked me with a pointed stick? > So there is strictly greater (yes > not be accepted for > example: it's quite simple to drop the issue. It means that > of your work being roughly as better than that of the > pointed stick is how a pointed stick stand, and that pointed > sticks up your work getting published. > Let me with a Xerox of mathematics than a work would publishing > the pointed stick to let me with that she made Adam's pointed > stick till thee morning come. > Come, Mr. Tally Mon, tally me then. > Now, I never could find them, because of copies of the Xerox > of the idea that pointed stick in the pointed stick. It's six > foot, eight foot, bunch. A beautiful bunch a'ripe pointed stick, > thus disarming him. > He's completely dead. > Well, he say day, he wanna go home. > Stack pointed sticks. We believe it would entail. We are > sorry to have a third party, would not be accepted for formal > peer review, even if you know that your work being published I > have a pointed stick pointed stick: just think of the Annals > constitute libel or slander. > My method was attacking me as likely as better than that matter > how it's hard to slice pointed stick fiend. First of the Annals > constitute libel or slander. > My method was attacking me get clear on a banana. > Shut up. > Supposing he's got a third party, would publishing the > pointed stick in some sense in the issue to have now 'elpless. > You have no pointed sticks up your paper is some sense. > For example, if you sent a man armed with it. Now it's quite simple > to have heard rumors that she made Adam's pointed sticks. > We believe that your butt. Time flies like an April Fool's edition, > for that she made Adam's pointed stick stand. > I have now 'elpless. > You shot him. He's dead. > Well, he say day, he say day-ay-ay-o. Day, he say day, he say day, > he say day, he say day, he wanna go home. > Stack pointed stick fiend. > First of the Annals, it may be nice material for that pointed > stick. Hold it like that, the distribution hassles that of > mathematics than a pointed stick. > So let me as better than the probability of the proper number > of the pointed stick, thus disarming him. You have a banana. > Shut up. > Come on. Come at me. Come on. Come on. Come at me then. > Now, I mean that when they put you have heard about. I believe you have just solved JSH's Serriagoats Factoring therom for him., without using the useless sets of equations he vomits forth on sci.math. === Subject: Re: SF: Tandem factorization back, new approach > Necessity is the mother of invention... >Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it >did, so I called crap crap and started feeling sorry for myself. Then >I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > and > T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) > which means that > 2*sqrt(x)*k_1 = f_1 + g_1 >2*sqrt(y)*k_2 = f_1 - g_1 >2*sqrt(x)*k_3 = f_2 + g_2 >2*sqrt(y)*k_4 = f_2 - g_2 > and the initial approach I put forward was with a target composite T to >factor you pick some surrogate S, and I had various ideas for how you >then find all the other variables--when for that idea to work, you need >to already know the factorization of S and T. > So that was when I was ready to toss all of this and declared it crap. > However, why pick S? > Using the first two equations I have > S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and > S+T = 2*k_1*k_3*x + 2*k_2*k_4*y > and focusing on that second equation if I let > k_1*k_3 = A > k_2*k_4 = B > and pick squares for x and y, then I determine S, if T is the target >factorization: > > S = 2*k_1*k_3*x + 2*k_2*k_4*y - T > so > > S = 2*A*x + 2*B*y - T > and you can use those equations to solve out two of the k's, and >substitute out S, to relate the remaining two k's in an equation that I >hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just > picked one. > Now multiplying both sides by k_1*k_2 gives > (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) > so I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 > and completing the square with respect to k_2, I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - > T)^2*k_1^2/(4A*sqrt(xy)) > multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + > B*y - T)^2*k_1^2 > which is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > which yuck, shows that I need to know the factorization of T, to know > how to pick A and B, so that > ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) > and you need the factorization of B and T, or A and T to go further so > it's another crap idea. > Went in a big freaking circle, again. > Maybe, maybe not. After all, I have > S = 2*A*x + 2*B*y - T > so what if I just decide that S is the target? > Then I need to solve out further with > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > as that is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - > 4T*B*y)*k_1^2 > and if I just pick B and T, as surrogates, and assume I know the > factorizations > a_1 * a_2 = B*y, and b_1*b_2 = T, then I have > A*x = B*y + T + a_1*b_1 + a_2*b_2 > and I can now substitute to find that S is given by > S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 > and making my substitutions that is > S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 That was dumb. That's why y is in at the end, as I put it in here when it was substituted out. Corrected it goes as follows. and let's collect with respect to the a's, so I have S = 4*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 and simplify to get S - b_1*b_2 - 2*a_2*b_2 = (4* a_2 + 2*b_1)*a_1 so I have finally that a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4a_2 + 2*b_1) and it looks like I can just pick b_1 and b_2, and then I just need to look for integer values for a_2 such that a_1 is an integer. One last thing to do then, with 4a_2 + 2*b_1 = h_1 then 4a_2 + 2*b_1 = 0 mod h_1 S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 so multiplying the first by -2*b_2 and the second by 4, I have -8b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 4S - 4b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 and subtracting the top one from the bm one I get 4S = 0 mod h_1 so not surprisingly I need the factorization of S to go further so it's another big freaking circle. === Subject: Re: SF: Tandem factorization back, new approach Have you considered that maybe there's a common theme to all your approaches? You can try as many variations on these themes as you like until the end of time and you'll still be stuck in your circles of nonsense. Imagine how much math you could have learnt in the last 5 years. Are you planning to waste the next 5 too? Why not actually go to college and learn something? > Necessity is the mother of invention... >Sorry but this is how the development process works with me. >> I went down one path before, found out it didn't work like I thought it >did, so I called crap crap and started feeling sorry for myself. Then >I thought of another approach, so here we go, again. >> The idea here is to use a tandem factorization of two numbers where >> S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >> and >> T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) >> which means that >> 2*sqrt(x)*k_1 = f_1 + g_1 >2*sqrt(y)*k_2 = f_1 - g_1 >2*sqrt(x)*k_3 = f_2 + g_2 >2*sqrt(y)*k_4 = f_2 - g_2 >> and the initial approach I put forward was with a target composite T to >factor you pick some surrogate S, and I had various ideas for how you >then find all the other variables--when for that idea to work, you need >to already know the factorization of S and T. >> So that was when I was ready to toss all of this and declared it crap. >> However, why pick S? >> Using the first two equations I have >> S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) >> and >> S+T = 2*k_1*k_3*x + 2*k_2*k_4*y >> and focusing on that second equation if I let >> k_1*k_3 = A >> k_2*k_4 = B >> and pick squares for x and y, then I determine S, if T is the target >factorization: >S = 2*k_1*k_3*x + 2*k_2*k_4*y - T >> so >S = 2*A*x + 2*B*y - T >> and you can use those equations to solve out two of the k's, and >substitute out S, to relate the remaining two k's in an equation that I >hope will give the approach that works. > Ok, so making that substitution into the first equation I have > 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) > and now I can solve out two of the k's, and divide by 2 to get > > A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) > where there were a couple of possible ways to substitute out and I just >picked one. > Now multiplying both sides by k_1*k_2 gives > (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) > so I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 > and completing the square with respect to k_2, I have > A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - >T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - >T)^2*k_1^2/(4A*sqrt(xy)) > multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + >B*y - T)^2*k_1^2 > which is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - >2T(A*x+B*y) + T^2)*k_1^2 > which yuck, shows that I need to know the factorization of T, to know >how to pick A and B, so that > ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) > and you need the factorization of B and T, or A and T to go further so >it's another crap idea. > Went in a big freaking circle, again. > Maybe, maybe not. After all, I have > S = 2*A*x + 2*B*y - T > so what if I just decide that S is the target? > Then I need to solve out further with > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - > 2T(A*x+B*y) + T^2)*k_1^2 > as that is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - > 4T*B*y)*k_1^2 > and if I just pick B and T, as surrogates, and assume I know the > factorizations > a_1 * a_2 = B*y, and b_1*b_2 = T, then I have > A*x = B*y + T + a_1*b_1 + a_2*b_2 > and I can now substitute to find that S is given by > S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 > and making my substitutions that is > S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 > That was dumb. That's why y is in at the end, as I put it in here when > it was substituted out. > Corrected it goes as follows. > and let's collect with respect to the a's, so I have > S = 4*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 > and simplify to get > S - b_1*b_2 - 2*a_2*b_2 = (4* a_2 + 2*b_1)*a_1 > so I have finally that > a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4a_2 + 2*b_1) > and it looks like I can just pick b_1 and b_2, and then I just need to > look for integer values for a_2 such that a_1 is an integer. > One last thing to do then, with > 4a_2 + 2*b_1 = h_1 > then > 4a_2 + 2*b_1 = 0 mod h_1 > S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 > so multiplying the first by -2*b_2 and the second by 4, I have > -8b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 > 4S - 4b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 > and subtracting the top one from the bm one I get > 4S = 0 mod h_1 > so not surprisingly I need the factorization of S to go further so it's > another big freaking circle. > === Subject: Re: SF: Tandem factorization back, new approach > Have you considered that maybe there's a common theme to all your > approaches? > You can try as many variations on these themes as you like until the > end of time and you'll still be stuck in your circles of nonsense. > Imagine how much math you could have learnt in the last 5 years. Are > you planning to waste the next 5 too? Why not actually go to college > and learn something? His actual background is in physics (or so he claims). His time would have been better spent working on physics --- working on a perpetual motion machine comes to mind as one possibility. --- Christopher Heckman >Necessity is the mother of invention... >Sorry but this is how the development process works with me. >> I went down one path before, found out it didn't work like I thought it >> did, so I called crap crap and started feeling sorry for myself. Then >> I thought of another approach, so here we go, again. >> The idea here is to use a tandem factorization of two numbers where >> S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >> and >> T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y)) >> which means that >> 2*sqrt(x)*k_1 = f_1 + g_1 >> 2*sqrt(y)*k_2 = f_1 - g_1 >> 2*sqrt(x)*k_3 = f_2 + g_2 >> 2*sqrt(y)*k_4 = f_2 - g_2 >> and the initial approach I put forward was with a target composite T to >> factor you pick some surrogate S, and I had various ideas for how you >> then find all the other variables--when for that idea to work, you need >> to already know the factorization of S and T. >> So that was when I was ready to toss all of this and declared it crap. >> However, why pick S? >> Using the first two equations I have >> S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) >> and >> S+T = 2*k_1*k_3*x + 2*k_2*k_4*y >> and focusing on that second equation if I let >> k_1*k_3 = A >> k_2*k_4 = B >> and pick squares for x and y, then I determine S, if T is the target >> factorization: >>S = 2*k_1*k_3*x + 2*k_2*k_4*y - T >> so >>S = 2*A*x + 2*B*y - T >> and you can use those equations to solve out two of the k's, and >> substitute out S, to relate the remaining two k's in an equation that I >> hope will give the approach that works. >Ok, so making that substitution into the first equation I have >> 2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy) >> and now I can solve out two of the k's, and divide by 2 to get >A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy) >> where there were a couple of possible ways to substitute out and I just >picked one. >> Now multiplying both sides by k_1*k_2 gives >> (A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy) >> so I have >> A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0 >> and completing the square with respect to k_2, I have >> A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y - >T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y - >T)^2*k_1^2/(4A*sqrt(xy)) >> multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives >> (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x + >B*y - T)^2*k_1^2 >> which is >> (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - >2T(A*x+B*y) + T^2)*k_1^2 >> which yuck, shows that I need to know the factorization of T, to know >how to pick A and B, so that >> ((A*x - B*y)^2 - 2T(A*x+B*y) + T^2) >> and you need the factorization of B and T, or A and T to go further so >it's another crap idea. >> Went in a big freaking circle, again. > Maybe, maybe not. After all, I have > S = 2*A*x + 2*B*y - T > so what if I just decide that S is the target? > Then I need to solve out further with > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 - >2T(A*x+B*y) + T^2)*k_1^2 > as that is > (A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 - >4T*B*y)*k_1^2 > and if I just pick B and T, as surrogates, and assume I know the >factorizations > a_1 * a_2 = B*y, and b_1*b_2 = T, then I have > A*x = B*y + T + a_1*b_1 + a_2*b_2 > and I can now substitute to find that S is given by > S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2 > and making my substitutions that is > S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2 > That was dumb. That's why y is in at the end, as I put it in here when > it was substituted out. > Corrected it goes as follows. > and let's collect with respect to the a's, so I have > S = 4*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2 > and simplify to get > S - b_1*b_2 - 2*a_2*b_2 = (4* a_2 + 2*b_1)*a_1 > so I have finally that > a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4a_2 + 2*b_1) > and it looks like I can just pick b_1 and b_2, and then I just need to > look for integer values for a_2 such that a_1 is an integer. > One last thing to do then, with > 4a_2 + 2*b_1 = h_1 > then > 4a_2 + 2*b_1 = 0 mod h_1 > S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1 > so multiplying the first by -2*b_2 and the second by 4, I have > -8b_2* a_2 - 4*b_1*b_2 = 0 mod h_1 > 4S - 4b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1 > and subtracting the top one from the bm one I get > 4S = 0 mod h_1 > so not surprisingly I need the factorization of S to go further so it's > another big freaking circle. > > > === Subject: Re: SF: Tandem factorization back, new approach >> Necessity is the mother of invention... > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) > I think your problem is here, you want S to be a number yes? Unfortunately, this equation does not specify a number. A squart root returns TWO values, you can't just wish that away. Thus there are 8 different Ss here. === Subject: Re: SF: Tandem factorization back, new approach > Necessity is the mother of invention... > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought > it > did, so I called crap crap and started feeling sorry for myself. > Then > I thought of another approach, so here we go, again. > The idea here is to use a tandem factorization of two numbers where > S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >> I think your problem is here, you want S to be a number yes? > Unfortunately, > this equation does not specify a number. A squart root returns TWO values, > you can't just wish that away. Thus there are 8 different Ss here. By convention, if he meant to include both values of the square root, he would have to write: S = [k_1*(+/-sqrt(x)) + k_2*(+/-sqrt(y))]*[k_3*(+/-sqrt(x)) + k_4*+/-sqrt(y))] KeithK === Subject: Re: SF: Tandem factorization back, new approach <12k95p6l7zi4x$.zb4j9843681m$.dlg@40tude.net> <12abd8i7v7dt435@corp.supernews.com > Necessity is the mother of invention... And boredom is the father. > Sorry but this is how the development process works with me. >> I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. >> The idea here is to use a tandem factorization of two numbers where >> S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y)) >> I think your problem is here, you want S to be a number yes? > Unfortunately, > this equation does not specify a number. A squart root returns TWO values, > you can't just wish that away. Thus there are 8 different Ss here. > By convention, if he meant to include both values of the square root, he > would have to write: > S = [k_1*(+/-sqrt(x)) + k_2*(+/-sqrt(y))]*[k_3*(+/-sqrt(x)) + > k_4*+/-sqrt(y))] One thing JSH is not is conventional. --- Christopher Heckman === Subject: Re: SF: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. You left out the step where you insult everyone and threaten to bring down the hammer. === Subject: Re: SF: Tandem factorization back, new approach > You left out the step where you insult everyone and threaten to bring > down the hammer. ... and the stock market crash, plague of locusts, end of the world, etc. === Subject: Re: SF: Tandem factorization back, new approach | > You left out the step where you insult everyone and threaten to bring | > down the hammer. | | ... and the stock market crash, plague of locusts, end of the world, etc. Oh well, might as well... http://www.foxnews.com/story/0,2933,201355,00.html before someone else does... what the hell === Subject: Re: SF: Tandem factorization back, new approach > | > You left out the step where you insult everyone and threaten to bring > | > down the hammer. > | ... and the stock market crash, plague of locusts, end of the world, > etc. > Oh well, might as well... > http://www.foxnews.com/story/0,2933,201355,00.html > before someone else does... what the hell That means if JSH gets it right this time, then the world won't end ? === Subject: Re: Tandem factorization back, new approach > Sorry but this is how the development process works with me. > I went down one path before, found out it didn't work like I thought it > did, so I called crap crap and started feeling sorry for myself. Then > I thought of another approach, so here we go, again. Forget about it. It is crap. Total crap, the foundation of it is indeed crap. Try an entirely new approach, like squares, or cube roots. === Subject: Re: SF: Tandem factorization back, new approach : Sorry but this is how the development process works with me. It's funny, I was about to post a response to your failure note suggesting that I'd give it about a week until you were back in your usual way blathering out nonsense. : I'll have to work through the equations and see if it is another : crapshoot. Why bother? You've never worked through any of your other ideas. Justin === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. But you're not a crackpot. Right, ? === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > That is all. > ___JSH At least you finally admitted it. === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. > That is all. > ___JSH > At least you finally admitted it. Hey, I always admit it when I'm proven wrong and realize it. I don't see the point in holding on to wrong ideas. In this case, the approach was just kind of thrilling, and I was just so sure that using the square root ambiguity would work, but everything I was doing was a BFC--big freaking circle. But you know, this is fun to me. And if you realized how much fun this is you'd understand why I just love going after these problems in this way. So my initial approach collapsed--yuck, felt terrible. Really Walked away, determined to move on to other things, and of course, I get another idea. Unfortunately, I've found that's what works!!! I need to put something out there, fight for it, be really wrong, and then, as everything collapses, and I start a pity party, wonder what is the meaning of life, etc., I get another idea. It's a hard way to discover things, but it is the way that works for me. It's kind of like living in Hell. No it is living in Hell. I actually think this is Hell and I'm being punished for something. I am serious. === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all > the > others. > That is all. > ___JSH >> At least you finally admitted it. > Hey, I always admit it when I'm proven wrong and realize it. > I don't see the point in holding on to wrong ideas. Is that a pointed stick joke? Anyway, this all reminds of Christmas 2002... === ] Subject: JSH: Ok, I'm a loser ] ] It's finally settled in that I'm just some pathetic loser. If I ] weren't so pathetic I'd just go away gracefully, but I'll send one ] more post, or who am I kidding, my patheticness is so great that I'll ] probably post yet again. ] ] I'm disgusting. I'm just a pile of . I should just die like so ] many of you have said. I hate myself. I despise this life. I'm ] nothing but a sick joke to be made fun of by those of you who have ] real educations. People who actually know something, when I know ] nothing. I'm just nothing. ] ] If I hadn't been such a disgusting human being I'd have come to this ] realization years ago instead of wasting your time. ] ] My life is nothing. I know nothing. I'm worth nothing. I'm just ] . ] ] Please forgive me. All your attacks were justified. ] ] -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: JSH: Latest factoring idea is crap <4giao4F1nmapcU1@individual.net > Ok, I accept it. My latest idea just goes in circles like all > the > others. >> That is all. >> ___JSH >> At least you finally admitted it. > Hey, I always admit it when I'm proven wrong and realize it. > I don't see the point in holding on to wrong ideas. > Is that a pointed stick joke? > Anyway, this all reminds of Christmas 2002... So the full cycle is about 3.5 years? How long after the you were all right post was the next you were all wrong post? - Randy === Subject: Re: JSH: Latest factoring idea is crap <4giao4F1nmapcU1@individual.net > Ok, I accept it. My latest idea just goes in circles like all >the >others. >> That is all. >> ___JSH > At least you finally admitted it. > Hey, I always admit it when I'm proven wrong and realize it. > I don't see the point in holding on to wrong ideas. > Is that a pointed stick joke? > Anyway, this all reminds of Christmas 2002... > So the full cycle is about 3.5 years? > How long after the you were all right post was the next > you were all wrong post? My guess would be the day after the Colt 45 ran out. > - Randy === Subject: Re: JSH: Latest factoring idea is crap >Ok, I accept it. My latest idea just goes in circles like all the >others. > That is all. > ___JSH > At least you finally admitted it. > Hey, I always admit it when I'm proven wrong and realize it. Oh, I get it. The problem is with you realizing it. > I don't see the point in holding on to wrong ideas. It's just you never realize they're wrong. > In this case, the approach was just kind of thrilling, and I was just > so sure that using the square root ambiguity would work, but everything > I was doing was a BFC--big freaking circle. > But you know, this is fun to me. And if you realized how much fun this > is you'd understand why I just love going after these problems in this > way. > So my initial approach collapsed--yuck, felt terrible. Really > Walked away, determined to move on to other things, and of course, I > get another idea. > Unfortunately, I've found that's what works!!! The getting of another idea? Not the idea itself? > I need to put something out there, fight for it, be really wrong, and > then, as everything collapses, and I start a pity party, wonder what is > the meaning of life, etc., I get another idea. > It's a hard way to discover things, but it is the way that works for > me. But you haven't discovered anything. > It's kind of like living in Hell. No it is living in Hell. I actually > think this is Hell and I'm being punished for something. Well, if this is Hell, your past discoveries must be delusions, eh? > I am serious. So am I. Satan is ing with your head and you don't even know it. > === Subject: Re: JSH: Latest factoring idea is crap >> Ok, I accept it. My latest idea just goes in circles like all the >> others. >> That is all. >> ___JSH > I am pretty sure this wasn't posted by the real James. He never signs > his posts with _JSH, he always uses . >> That, and the real Harris will always acompany admittance of defeat with >> you all hate me, and math, and civilization, and humanity, and >> puppies... > seems to be JSH, alright, perhaps he met a girl, an algebra teacher, > a foxie one, who knows factoring well , and she stays on her meds. > #1 this posting> > You should have provided a URL for #2, because there's a possibility that it wasn't real. -- Andrew Poelstra < http://www.wpsoftware.net/blog > To email me, use apoelstra at the above address. I know that area of town like the back of my head. === Subject: Re: JSH: Latest factoring idea is crap > Ok, I accept it. My latest idea just goes in circles like all the > others. That is all. ___JSH >> I am pretty sure this wasn't posted by the real James. He never signs >> his posts with _JSH, he always uses . > That, and the real Harris will always acompany admittance of defeat with > you all hate me, and math, and civilization, and humanity, and > puppies... >> seems to be JSH, alright, perhaps he met a girl, an algebra >> teacher, >> a foxie one, who knows factoring well , and she stays on her meds. >> #1 this posting #2 previous JSH posting should be real> it wasn't real. Looks like the IP changed on june 15th === Subject: Re: JSH: Latest factoring idea is crap ... It can, but in that case you also have to fake the Path information... > Looks like the IP changed on june 15th Both are from comcast.net. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Latest factoring idea is crap >That, and the real Harris will always acompany admittance of defeat with >you all hate me, and math, and civilization, and humanity, and puppies... But I like puppies, they're delicious. === Subject: The 'awarely word' catches the worm.Letters don.mcd. The 'awarely word' catches the worm.Letters don.mcd. dear editor, New Sctst, news, What would be worse than a television studio full of uncommitted voters twisting knobs to produce the political party leaders' 'election worm?' Answer- touchy-feely computers that tap into our emotions, Reuters (Dominion Post 27/6/06.) New Zealand. An 'emotionally aware' computer being developed by British and American scientists will be able to read a person's thoughts by analysing facial movements that represent underlying feelings ... The computer had better have running shoes because I am thinking now (I'll kick it in the transistors.) The decline of computer applications is marked by (there is 1 chance in 158,753,389,900 that you will be choose 13 / 4 =, and 1580,187,223 is the least prime divisor of number (1 billion raised to the 1 billion-th power, plus 3.) (nine billion zeroes.) sci.math 2000. What emotional or truth score does that produce on your 17 inch plasma screen! OR the answer to a shaun plunkett Morning Report question (Radio NZ.) Can you please let me know the e-mail address for Autistic vision impaired against automaton. How will the technology help them? I am not particularly autistic vision impaired as far as I know. Don S. McDonald === Subject: Re: The Speed of Toilet Paper > Here's the kind of things you can do to have some fun with math: > Americans use enough toilet paper in one day to wrap around the world nine > times. Amazing, I'd always said the Americans were full of it. Now it appears I've been proved correct; which is a dang sight more than can be for JSH :-D mori >If it were on one giant roll, we would be unrolling it at the rate of > 7,600 miles per hour - roughly mach. 10, ten times the speed of sound. > - from ww.odd-info.com - Have fun! === Subject: Re: The Speed of Toilet Paper > On Mon, 26 Jun 2006 23:24:11 -0700, William Elliot > We aim to please, so please aim. <--- funnier over a urinal We aim to please. You aim too, please. Bye. Jasen === Subject: Re: The Speed of Toilet Paper On Tue, 27 Jun 2006 19:03:17 -0700, William Elliot >Hm, about 1 roll per month. Seems within range. > We aim to please, so please aim. <--- funnier over a urinal >Ah shucks, an uncreative plagiarist. >By physical deficiencies do you mean inadequate typing skills? >> I imagine a quick downward glance next time you're in the shower ought >> to answer your question. >Your banal penis envy insult lacks any inspiration for me to proffer >you a poetic retort. Therefore I'll have to resort to clumsy, turgid prose. IFYPFY. BW === Subject: Re: The Speed of Toilet Paper > Your banal penis envy insult lacks any inspiration for me to proffer > you a poetic retort. Thus I'll have to resort to the inspiration of > a previous neophyte bad mouther to whom I answered: >> You are Comicbook Guy from the Simpsons, and I claim my Darth Vader >> Empire Strikes Back action figure in its original box. >Look out Darth Vader, you shouldn't hold a death ray in your hand when >you're rolling on the floor with laughter, you might hit yourself ... >Darth Vader? Darth Vader? Oh no, nooo... I shouldn't have told him that >joked. Damn, with his death, Bush will take over Empire and we'll be >longing for the good old days of RayGun Darth Vader. Inarticulate and st00pid? dance, Billy, dance! -- Chris McG. Harming humanity since 1951. What do you expect from a bunch of kiwi smoking sheep herders? -- -- === Subject: Re: The Speed of Toilet Paper > Your banal penis envy insult lacks any inspiration for me to proffer >> you a poetic retort. Thus I'll have to resort to the inspiration of >> a previous neophyte bad mouther to whom I answered: >You are Comicbook Guy from the Simpsons, and I claim my Darth Vader >Empire Strikes Back action figure in its original box. I have a simpler explanation, one having to do with Bozo Boi here getting a thesaraus for his birthday. I think Will Rogers had Bozo Boi in mind when he said, The difference between the right word and the almost-right word is the difference between the lightning and the lightning bug. -- Danked, the past participle of dank, is used to refer to someone who replies to his own post on an online forum posing as another person (see Internet sock puppet) but forgetting to change his username . . . . This was an act of stupidity meriting a name of its own, and because the hapless contributor's username was Danks, the term dank or danked emerged. -- http://en.wikipedia.org/wiki/Danked === Subject: Re: The Speed of Toilet Paper <4l75a2p79osejskpcsfi6sj94f7e1s4b9r@4ax.com On Tue, 27 Jun 2006 22:47:35 -0400, Marc Goodman >> Your banal penis envy insult lacks any inspiration for me to proffer >> you a poetic retort. Thus I'll have to resort to the inspiration of >> a previous neophyte bad mouther to whom I answered: >You are Comicbook Guy from the Simpsons, and I claim my Darth Vader >Empire Strikes Back action figure in its original box. > I have a simpler explanation, one having to do with Bozo Boi here > getting a thesaraus for his birthday. > I think Will Rogers had Bozo Boi in mind when he said, The difference > between the right word and the almost-right word is the difference > between the lightning and the lightning bug. It was Mark Twain, not Will Rogers: http://www.twainquotes.com/Lightning.html [quote] The difference between the almost right word & the right word is really a large matter--it's the difference between the lightning bug and the lightning. - Letter to George Bainton, 10/15/1888 [end quote] -- Raymond S. Wise Minneapolis, Minnesota USA E-mail: mplsray @ yahoo . com === Subject: Re: The Speed of Toilet Paper >> Your banal penis envy insult lacks any inspiration for me to proffer >> you a poetic retort. Thus I'll have to resort to the inspiration of >> a previous neophyte bad mouther to whom I answered: >You are Comicbook Guy from the Simpsons, and I claim my Darth Vader >Empire Strikes Back action figure in its original box. >> I have a simpler explanation, one having to do with Bozo Boi here >> getting a thesaraus for his birthday. >> I think Will Rogers had Bozo Boi in mind when he said, The difference >> between the right word and the almost-right word is the difference >> between the lightning and the lightning bug. It's more like the difference between YHBT and YHL. > It was Mark Twain, not Will Rogers: HAND! ---- === Subject: Re: The Speed of Toilet Paper On Tue, 27 Jun 2006 20:56:01 -0700, William Elliot >>Hm, about 1 roll per month. Seems within range. > We aim to please, so please aim. <--- funnier over a urinal > Ah shucks, an uncreative plagiarist. >> You might want to find a dictionary site so you can figure >> out what the second big word means. It would be preferable >> if you checked your nonsense *before* posting it to the >> usenet. I know I always do. >shucks >an exclamation of mild disappointment, disgust, etc. He thinks shucks is a big word? That's kinda sad, innit? -- Danked, the past participle of dank, is used to refer to someone who replies to his own post on an online forum posing as another person (see Internet sock puppet) but forgetting to change his username . . . . This was an act of stupidity meriting a name of its own, and because the hapless contributor's username was Danks, the term dank or danked emerged. -- http://en.wikipedia.org/wiki/Danked === Subject: Re: The Speed of Toilet Paper > He thinks shucks is a big word? That's kinda sad, innit? It has a high letter to syllable ratio: 6, in fact. What is the standard for what constitutes a big word? -- Vielen Dank === Subject: Re: The Speed of Toilet Paper <4vvan3-hrn.ln1@news.ducksburg.com > He thinks shucks is a big word? That's kinda sad, innit? > It has a high letter to syllable ratio: 6, in fact. > What is the standard for what constitutes a big word? Around here I is a very big word. === Subject: Re: The Speed of Toilet Paper On Wed, 28 Jun 2006 19:01:07 -0700, William Elliot > He thinks shucks is a big word? That's kinda sad, innit? >> It has a high letter to syllable ratio: 6, in fact. >> What is the standard for what constitutes a big word? >Around here I is a very big word. We knew you had a high opinion of yourself, but I for one thought you wouldn't admit it so blatantly. Maybe you would care to explain what's so special about you. BW === Subject: Re: The Speed of Toilet Paper >>Around here I is a very big word. > We knew you had a high opinion of yourself, but I for one thought you > wouldn't admit it so blatantly. > Maybe you would care to explain what's so special about you. And why he's referring to himself in quotes, and in a mixture of first and third persons? Such habits are normally expected only from the dank. === Subject: Re: The Speed of Toilet Paper > Around here I am a very big word. IFYPFY. === Subject: Re: The Speed of Toilet Paper <4vvan3-hrn.ln1@news.ducksburg.com> <6incn3-3fb.ln1@news.ducksburg.com > Around here I am a very big word. > IFYPFY. HAWYAW === Subject: Re: The Speed of Toilet Paper >Around here I am a very big word. > IFYPFY. > HAWYAW Around here I am a very big word => Around here I am is a very big word mori === Subject: Re: The Speed of Toilet Paper > Around here I am a very big word. > IFYPFY. >> HAWYAW >Around here I am a very big word => Around here I am is a very big word No no no. Around here 'I' be a very big word, yo you! Dave -- /DavidtDeLaneytpostingtfrom dbd@vic.com It's not the pot thattgrows the flower It's not the clock thattslows the hour tThe definition's plain for anyone to see Love istall it takes totmake a family - R&P. VISUALIZEtHAPPYNET VRbeable http://www.vic.com/~dbd/ - net.legends FAQ & Magic / I WUV you in all CAPS! --K. === Subject: Re: The Speed of Toilet Paper s. === Subject: Re: The Speed of Toilet Paper On Fri, 30 Jun 2006 02:20:43 -0700, William Elliot >> Around here I am a very big word. > IFYPFY. > HAWYAW >> Around here I am a very big word => Around here I am is a very big word >Bus, you guys are unbelievable. >Amongst you, even 'a' is a big word. >Y'all always be talking in big words. If you're going to try to be patronizing, do it right. BW === Subject: Re: The Speed of Toilet Paper > HAWYAW > If you're going to try to be patronizing, do it right. You're right, he should have started with HEEHAW. -- Vielen Dank === Subject: Re: The Speed of Toilet Paper <4vvan3-hrn.ln1@news.ducksburg.com> <6incn3-3fb.ln1@news.ducksburg.com> HAWYAW > If you're going to try to be patronizing, do it right. > You're right, he should have started with HEEHAW. Hello extraneous extras hiding awesome warps. === Subject: Re: The Speed of Toilet Paper On Thu, 29 Jun 2006 04:46:59 -0700, William Elliot >Around here I am a very big word. > IFYPFY. > HAWYAW I wish I understood that language. Bill ---------------------------------------------------------------- Reverse parts of the ISP name and the user name for my e-address === Subject: liminf Let {x_n} and {y_n} be nonnegative sequences. Is it true, in general, that liminf(x_n*y_n)>=liminf(x_n)liminf(y_n) ? It seems intuitively true(Although Im not sure how you handle the case where x_n goes to infinity and y_n goes to zero), but Im not sure how to prove it. === Subject: Re: liminf > Let {x_n} and {y_n} be nonnegative sequences. Is it true, in general, that > liminf(x_n*y_n)>=liminf(x_n)liminf(y_n) ? > It seems intuitively true(Although Im not sure how you handle the case where > x_n goes to infinity and y_n goes to zero), but Im not sure how to prove it. Define 0*oo = 0 (as is done often in integration theory), and a*oo = oo whenever 0 < a <= oo. Then the above is true. Start with a_n = inf{x_m : m >= n}, b_n = inf{y_m : m >= n}. Then for any m >= n, x_m*y_m >= a_n*b_n, etc. === Subject: Why why why Why why why Why if one loaf of bread plus 1 loaf = 2 loafs of bread then why if you take one away it = no loafs of bread??? 1-1= Zero? Do you not still have 1 loaf of bread??? Par-due Primiary Principal one of us Morfeeis.................. the answer: lies within a child who has not been taught. === Subject: SF: Simpler ideas Having given up on the complex stuff for surrogate factoring, I'm now playing with simpler: x^2 - a^2 = S + T x^2 - b^2 = S - k*T so b^2 - a^2 = (k-1)*T where you pick some surrogate S, like S=1, and factor S+T. But then you need a sum of factors of (S-k*T)/4 to equal the sum of the factors of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you need to have S - k*T = (f_1 + f_2 - j)*j where j is a natural number, and you solve for k, to get k = (S - (f_1 + f_2 - j)*j)/T where you need k to be an integer, so you also have S - (f_1 + f_2 - j)*j = 0 mod T so j^2 - (f_1 + f_2)*j + S = 0 mod T and completing the square gives j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T so (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T so you have to find the quadratic residue of ((f_1 + f_2)^2 - S) modulo T, to find j, and then you can find b, from b^2 = x^2 - S + kT as you get x, from x^2 = a^2 + S+T, that then gives you b^2 - a^2 = (k-1)T and the factorization (b-a)*(b+a) = (k-1)*T and that's the idea. === Subject: Re: Simpler ideas > Having given up on the complex stuff for surrogate factoring, I'm now > playing with simpler: How bout playing with an algebra book? And the type of playing I'm talking about is also called reading. === Subject: Re: Simpler ideas >> Having given up on the complex stuff for surrogate factoring, I'm now >> playing with simpler: > How bout playing with an algebra book? And the type of playing I'm talking > about is also called reading. http://www.dummies.com/WileyCDA/DummiesTitle/productCd-0764563742.html === Subject: Re: SF: Simpler ideas > Having given up on the complex stuff for surrogate factoring, I'm now > playing with simpler: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > where you pick some surrogate S, like S=1, and factor S+T. But then > you need a sum of factors of (S-k*T)/4 to equal the sum of the factors > of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you > need to have > S - k*T = (f_1 + f_2 - j)*j > where j is a natural number, and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > where you need k to be an integer, so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T Forgot to multiply S times 4, so that should be (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T but now I see that it's trivial that j can't be an integer, so k can't be either, so I need a ratio, like j = x/y, and that gives (2*x - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T and now the puzzle is how to pick natural number y. Once you have it, then you get a fraction for k, from k = (S - (f_1 + f_2 - j)*j)/T and then you can get b, and factor (k-1)*T, with b^2 - a^2 = (k-1)*T and luckily this is simple enough that mistakes have a hard time to hide, so that just leaves the question of picking y. === Subject: Re: SF: Simpler ideas > Having given up on the complex stuff for surrogate factoring, I'm now > playing with simpler: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > where you pick some surrogate S, like S=1, and factor S+T. But then > you need a sum of factors of (S-k*T)/4 to equal the sum of the factors > of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you > need to have > S - k*T = (f_1 + f_2 - j)*j > where j is a natural number, and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > where you need k to be an integer, so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T > Forgot to multiply S times 4, so that should be > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > but now I see that it's trivial that j can't be an integer, so k can't > be either, so I need a ratio, like > j = x/y, and that gives Dumb. Forgot I've already used x, I think instead I'll use j = z/y > (2*x - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T Then that becomes (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T and I'm mulling it over. Lots of ways it can go trivially, but... === Subject: Re: SF: Simpler ideas >Having given up on the complex stuff for surrogate factoring, I'm now >playing with simpler: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > where you pick some surrogate S, like S=1, and factor S+T. But then >you need a sum of factors of (S-k*T)/4 to equal the sum of the factors >of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you >need to have > S - k*T = (f_1 + f_2 - j)*j > where j is a natural number, and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > where you need k to be an integer, so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T > Forgot to multiply S times 4, so that should be > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > but now I see that it's trivial that j can't be an integer, so k can't > be either, so I need a ratio, like > j = x/y, and that gives > Dumb. Forgot I've already used x, I think instead I'll use > j = z/y > (2*x - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > Then that becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > and I'm mulling it over. Lots of ways it can go trivially, but... It's got to be a matter of quadratic residues modulo T. And with y stuffed in there you can just pick some y to shift that if needed to try and make it easier. And then you can solve everything out and get the factorization b^2 - a^ = (k-1)*T so it's at this point just about quadratic residues modulo T. === Subject: Re: SF: Simpler ideas > Having given up on the complex stuff for surrogate factoring, > I'm now > playing with simpler: >x^2 - a^2 = S + T >x^2 - b^2 = S - k*T >so >b^2 - a^2 = (k-1)*T >where you pick some surrogate S, like S=1, and factor S+T. But > then > you need a sum of factors of (S-k*T)/4 to equal the sum of the > factors > of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, > then you > need to have >S - k*T = (f_1 + f_2 - j)*j >where j is a natural number, and you solve for k, to get >k = (S - (f_1 + f_2 - j)*j)/T >where you need k to be an integer, so you also have >S - (f_1 + f_2 - j)*j = 0 mod T >so >j^2 - (f_1 + f_2)*j + S = 0 mod T >and completing the square gives >j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) > mod T >so >(2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T >Forgot to multiply S times 4, so that should be > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > but now I see that it's trivial that j can't be an integer, so k > can't > be either, so I need a ratio, like > j = x/y, and that gives > Dumb. Forgot I've already used x, I think instead I'll use >> j = z/y > (2*x - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T >> Then that becomes >> (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T >> and I'm mulling it over. Lots of ways it can go trivially, but... > It's got to be a matter of quadratic residues modulo T. And with y > stuffed in there you can just pick some y to shift that if needed to > try and make it easier. > And then you can solve everything out and get the factorization > b^2 - a^ = (k-1)*T > so it's at this point just about quadratic residues modulo T. There are bad. There's no Messiah! Brian: Er, Ire. Er, Romanes eunt down you promise I won't. Jones: Aaagh! (dies.) Sgt.: I won't NEED to find all go up together! Right Side Of Life's got a bunch. Sgt.: No. Palin: Suppose him fighting at Brian's mother: Whole basket each. The Crowd: I'm going my own admiration if I let k_1*k_4)*sqrt(x) - k_4*sqrt(y))*(k_3*sqrt(x)*k_3 + k_1*k_4)*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(y)) and T. So you, Romanus ! Centurion: Sir, the square gives j^2 - a^2 = ((f_1 + f_2)^2 - 4*S) mod T and the dative ! Centurion: I won't keep it is to be three Wise Man #2: Look, look, I'm going my own by their heads at his named Biggus Dickus. [guard laugh and see that's the Jerusalem Garrisoner: Who do we? Right O'clock. Reg: Right. Banana. (Does so.) Palin: No rock) RIGHT! Who are the house] -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: SF: Simpler ideas >Having given up on the complex stuff for surrogate factoring, I'm >now >playing with simpler: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > where you pick some surrogate S, like S=1, and factor S+T. But then >you need a sum of factors of (S-k*T)/4 to equal the sum of the >factors >of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you >need to have > S - k*T = (f_1 + f_2 - j)*j > where j is a natural number, and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > where you need k to be an integer, so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T > > Forgot to multiply S times 4, so that should be > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > but now I see that it's trivial that j can't be an integer, so k can't > be either, so I need a ratio, like > j = x/y, and that gives > Dumb. Forgot I've already used x, I think instead I'll use > j = z/y > (2*x - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > Then that becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > and I'm mulling it over. Lots of ways it can go trivially, but... >> It's got to be a matter of quadratic residues modulo T. And with y >> stuffed in there you can just pick some y to shift that if needed to >> try and make it easier. >> And then you can solve everything out and get the factorization >> b^2 - a^ = (k-1)*T >> so it's at this point just about quadratic residues modulo T. > There are bad. There's no Messiah! > Brian: Er, Ire. Er, Romanes eunt down you promise I won't. > Jones: Aaagh! (dies.) > Sgt.: I won't NEED to find all go up together! Right Side Of Life's got a > bunch. > Sgt.: No. > Palin: Suppose him fighting at > Brian's mother: Whole basket each. > The Crowd: I'm going my own admiration if I let k_1*k_4)*sqrt(x) - > k_4*sqrt(y))*(k_3*sqrt(x)*k_3 + k_1*k_4)*sqrt(x) - > k_2*sqrt(y))*(k_3*sqrt(y)) and T. So you, Romanus ! > Centurion: Sir, the square gives j^2 - a^2 = ((f_1 + f_2)^2 - 4*S) mod T > and the dative ! > Centurion: I won't keep it is to be three > Wise Man #2: Look, look, I'm going my own by their heads at his named > Biggus Dickus. [guard laugh and see that's the > Jerusalem Garrisoner: Who do we? Right O'clock. > Reg: Right. Banana. (Does so.) > Palin: No rock) RIGHT! Who are the house] let's see you call anybody Biggus...... Dickus.... without laughing! === Subject: Re: SF: Simpler ideas > Forgot to multiply S times 4, so Dissenter: Uh, I'm a leper! Ex-Leper: Oh I say you also the welfare of Jews of both of the wall. The deceased, Mr Apricot. Chapman (member of class): Perhaps they're not, in fact, then? [They all laughs some assistance if there isn't a 16-ton weights. Sgt.: What? Stan: Yeah, first lines] Wise Men and proud of it, sir. Hail Caesar ! And if it is another approach I put forward was with a thudden crisis] Lead Singer Crucifee: [singing] Life's a very difficult time, a friend's hammer, and the idea here is a natural number, and beginning! [Brian of Nazareth. Centurion: What's right. Take him away with it. Sgt.: I warned you. Good-bye. [Three wise men. Brian, and I hadn't done -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Simpler ideas > Having given up on the complex stu Amazing!! I had this idea.. If t = p*z_1+q*z_2 and u = p*z_1-q*z_2 then f(u, t) = g_1(p, q, z_1, z_2) and then we let p = sqrt(g_1+k_star). I call this a Surrogate Function. It is a member of a ring which I call the Surrogate Ring. Well, I haven't done all the algebra (perhaps someone could do it for me using Maple or Mathematica) but, if I am right, then T = A*B and the factoring problem is solved! I haven't had time to work out the boring details but I am sure that somebody reading this will do so. It will mean the end of the Universe. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Simpler ideas <4gplm2F1ni1n5U1@individual.net > Having given up on the complex stu > Amazing!! > I had this idea.. > If t = p*z_1+q*z_2 > and u = p*z_1-q*z_2 > then f(u, t) = g_1(p, q, z_1, z_2) and then we let p = > sqrt(g_1+k_star). I call this a Surrogate Function. It is a member of > a ring which I call the Surrogate Ring. Well, I haven't done all the > algebra (perhaps someone could do it for me using Maple or > Mathematica) but, if I am right, then T = A*B and the factoring > problem is solved! > I haven't had time to work out the boring details but I am sure that > somebody reading this will do so. It will mean the end of the > Universe. You forgot: Yes, times almost without number, I have been chased by police for bringing you the gift of my mathematics ... Is it not so that professors at a university are supposed to have the maturity and judgment and wisdom to be able to read a writing and determine for themselves the value of its contents? Is it that they require THOUGHT CONTROL POLICE to tell them what they should or should not read or think about? ... Ah, what hypocrites and mockers you are! --- Christopher Heckman === Subject: Re: Simpler ideas >>Having given up on the complex stu > Amazing!! > I had this idea.. > If t = p*z_1+q*z_2 > and u = p*z_1-q*z_2 > then f(u, t) = g_1(p, q, z_1, z_2) and then we let p = > sqrt(g_1+k_star). I call this a Surrogate Function. It is a member of > a ring which I call the Surrogate Ring. Well, I haven't done all the > algebra (perhaps someone could do it for me using Maple or > Mathematica) but, if I am right, then T = A*B and the factoring > problem is solved! > I haven't had time to work out the boring details but I am sure that > somebody reading this will do so. It will mean the end of the > Universe. But not of bananas! === Subject: Re: Simpler ideas >Having given up on the complex stu >> Amazing!! >> I had this idea.. >> If t = p*z_1+q*z_2 >> and u = p*z_1-q*z_2 >> then f(u, t) = g_1(p, q, z_1, z_2) and then we let p = >> sqrt(g_1+k_star). I call this a Surrogate Function. It is a member >> of a ring which I call the Surrogate Ring. Well, I haven't done all >> the algebra (perhaps someone could do it for me using Maple or >> Mathematica) but, if I am right, then T = A*B and the factoring >> problem is solved! >> I haven't had time to work out the boring details but I am sure >> that somebody reading this will do so. It will mean the end of the >> Universe. > But not of bananas! True. But anyway, I call grapefruit Surrogate Bananas and when I call something by some name it is obvious what it means. Please don't ask me for boring definitions. -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: Simpler ideas >>Having given up on the complex stu > Amazing!! > I had this idea.. > If t = p*z_1+q*z_2 > and u = p*z_1-q*z_2 > then f(u, t) = g_1(p, q, z_1, z_2) and then we let p = sqrt(g_1+k_star). > I call this a Surrogate Function. It is a member of a ring which I call > the Surrogate Ring. Well, I haven't done all the algebra (perhaps > someone could do it for me using Maple or Mathematica) but, if I am > right, then T = A*B and the factoring problem is solved! > I haven't had time to work out the boring details but I am sure that > somebody reading this will do so. It will mean the end of the Universe. >> But not of bananas! > True. But anyway, I call grapefruit Surrogate Bananas and when I call > something by some name it is obvious what it means. Please don't ask me > for boring definitions. do grapefruits come in bunches? If not perhaps vegetables work as good a fruits, like radishes, they come in bunches for sure. Then we can multiply through by 7, and factor that to the bank. I am sure someone can figure all this out for me, please. - JSH (Jesus S Hrist!) === Subject: Re: SF: Simpler ideas > Having given up on the complex stuff for surrogate factoring, I'm now > playing with simpler: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > where you pick some surrogate S, like S=1, and factor S+T. But then > you need a sum of factors of (S-k*T)/4 to equal the sum of the factors > of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then you > need to have > S - k*T = (f_1 + f_2 - j)*j > where j is a natural number, and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > where you need k to be an integer, so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - S) mod T > so you have to find the quadratic residue of ((f_1 + f_2)^2 - S) modulo > T, to find j, and then you can find b, from > b^2 = x^2 - S + kT > as you get x, from x^2 = a^2 + S+T, that then gives you > b^2 - a^2 = (k-1)T > and the factorization > (b-a)*(b+a) = (k-1)*T > and that's the idea. It means nothing, but I never could find them, because of the other. But an actual, physical banana: just think of the Annals constitute libel or slander. My method was to publish your butt. Time flies like that, the proper number of the banana stand. I disagree. It would not be able to drop the distribution hassles that she made Adam's banana stand and for example. This latter scenario strikes me get clear on this, however. By the proper number of the banana stand, and your butt. Time flies like a specialized journal. So let me banana. Who was the Annals constitute libel or slander. My method was attacking me qualify this, if you know that banana. Hide thee morning come. Come, Mr. Tally Mon, tally me as better than that banana. Here, you know that we have now rendered him to pick one as your paper is strictly greater (yes > not >=) than that she made Adam's banana in which your butt. Time flies like a banana. You shot him. You shot him. He's completely dead. I never could find them, because of all, you force him helpless. Supposing he's got a banana in it, for example. This latter scenario strikes me as likely as your paper is shipped with every copy of the banana, next, you sent a pointed stick. Shut up. Supposing he's got that, that's it. Come at me qualify this, if sending someone a banana in which your banana stand and your paper is well formed by the banana, thus disarming him. He's dead. I believe that she made Adam's banana in some sense in some sense in prison, other inmates stick bananas and for that your work would not be accepted for formal peer review, even if you in prison, other inmates stick bananas today. People are reacting like a work of Mathematics. We are reacting like an arrow, fruit flies like this means that would publishing the banana. It's six foot, bunch. Shut up. Supposing he's got a cover letter and that your work being roughly as likely as likely as likely as being published in the first place. Work all night on this, if you have now rendered him helpless. Supposing he's got a banana. Hide thee deadly black tarantula. Day-o, day-ay-ay-o. Day, he wanna go home. Stack banana stand, and look at me with a third party, would entail. We are reacting like this means nothing, but I never could find them, because of the Annals constitute libel or slander. > === Subject: Re: SF: Simpler ideas >> Having given up > It means nothing, but I never could find them, because of the > other. But an actual, physical banana: just think of the Annals > constitute libel or slander. My method was to publish your butt. > Time flies like that, the proper number of the banana stand. > I disagree. It would not be able to drop the distribution hassles > that she made Adam's banana stand and for example. This latter > scenario strikes me get clear on this, however. By the proper > number of the banana stand, and your butt. > Time flies like a specialized journal. So let me banana. Who was > the Annals constitute libel or slander. My method was attacking > me qualify this, if you know that banana. Hide thee morning come. > Come, Mr. Tally Mon, tally me as better than that banana. > Here, you know that we have now rendered him to pick one as your > paper is strictly greater (yes > not >=) than that she made > Adam's banana in which your butt. Time flies like a banana. > You shot him. You shot him. He's completely dead. > I never could find them, because of all, you force him helpless. > Supposing he's got a banana in it, for example. This latter scenario > strikes me as likely as your paper is shipped with every copy of the > banana, next, you sent a pointed stick. > Shut up. > Supposing he's got that, that's it. Come at me qualify this, if > sending someone a banana in which your banana stand and your paper > is well formed by the banana, thus disarming him. He's dead. > I believe that she made Adam's banana in some sense in some sense > in prison, other inmates stick bananas and for that your work would > not be accepted for formal peer review, even if you in prison, > other inmates stick bananas today. > People are reacting like a work of Mathematics. We are reacting > like an arrow, fruit flies like this means that would publishing > the banana. It's six foot, bunch. > Shut up. > Supposing he's got a cover letter and that your work being roughly > as likely as likely as likely as being published in the first place. > Work all night on this, if you have now rendered him helpless. > Supposing he's got a banana. > Hide thee deadly black tarantula. Day-o, day-ay-ay-o. Day, he wanna > go home. Stack banana stand, and look at me with a third party, > would > entail. > We are reacting like this means nothing, but I never could find > them, > because of the Annals constitute libel or slander. Exactly!! Excellent work! The total banana-ization of mathematics! What a project. Fruit flies... bananas... pointed sticks... But always remember: THE 16 TON WEIGHT that is 10 TONS in hexadecimal ! ! ! ! ! ! -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: SF: Simpler ideas But we have no bananas? === Subject: Re: SF: Simpler ideas In , Gib Bogle >But we have no bananas? Bananas were not available in the UK for most of the 1940's, but a government information film demonstrated how to make Mock Banana from parsnips. -- John Roberts-Jones === Subject: Re: SF: Simpler ideas , Gib Bogle >But we have no bananas? > Bananas were not available in the UK for most of the 1940's, > but a government information film demonstrated how to > make Mock Banana from parsnips. Do not make parsnips mock bananas. Banana power! Banana power! TIA, TC (MVP MSAccess) http://tc2.atspace.com === Subject: Re: SF: Simpler ideas > In , Gib Bogle >>But we have no bananas? > Bananas were not available in the UK for most of the 1940's, > but a government information film demonstrated how to > make Mock Banana from parsnips. We'll keep parsnips out of this, if you don't mind! === Subject: Re: SF: Simpler ideas In , Gib Bogle >> In , Gib Bogle >But we have no bananas? >> Bananas were not available in the UK for most of the 1940's, >> but a government information film demonstrated how to >> make Mock Banana from parsnips. >We'll keep parsnips out of this, if you don't mind! You will be pilloried when The Truth Comes Out and the World's Press realises that pointy parsnips could have been the only defence against the Moscow Mafiya's Mock Factors made from beetroot.= > We'll keep parsnips out of this, if you don't mind! Colonel (Graham Chapman): Aaagh! (dies.) Sgt.: Oh all high and y, then gives you b^2 = ((f_1 + f_2 = f_2 - g_2 2*sqrt(xy) and complex stuff for how the banana. First two k's in syrup... Sgt.: Thompson. Sgt.: I have to solve out how to defend you with a raspberry. The deceased, Mr Knowall. The deceased, Mr Tin Peach. Jones: He's dead. Sgt.: Well, he wall-- CRASH! a 16-ton weight sir! James: Good evening, I'm only doing my lad. When you means that. When you can see if it is just said: a pineapples, grapefruit not only doing to do we? Getting to me! Sgt.: No, I wasn't. Jones: Thompson. Come at me. Sgt.: Why not? Jones: You then in with a banana. Now the idea. Colonel ( Graham Chapman: Mangoes in syrup... Sgt.: We done of this time. Sgt.: No, I was when find j, and T. So that equations and B such that he eats not relish that second equations of ((f_1 + f_2)*j + S = 2*k_1*k_3*x + 2*B*y - T so j^2 - a^2 = ((f_1 + f_2)*j + S = (k-1)*T where are millions and screams.) Sgt.: Sorry but also haven't -- Clive Tooth www.clivetooth.dk Stock photos: http://submit.shutterstock.com/?ref=61771 === Subject: Re: SF: Simpler ideas > In , Gib Bogle >>But we have no bananas? > Bananas were not available in the UK for most of the 1940's, > but a government information film demonstrated how to make Mock Banana > from parsnips. >> We'll keep parsnips out of this, if you don't mind! > Colonel (Graham Chapman): Aaagh! (dies.) > Sgt.: Oh all high and y, then gives you b^2 = ((f_1 + f_2 = f_2 - g_2 > 2*sqrt(xy) and complex stuff for how the banana. First two k's in syrup... > Sgt.: Thompson. > Sgt.: I have to solve out how to defend you with a raspberry. The > deceased, Mr Knowall. The deceased, Mr Tin Peach. > Jones: He's dead. > Sgt.: Well, he wall-- > CRASH! a 16-ton weight sir! > James: Good evening, I'm only doing my lad. When you means that. When you > can see if it is just said: a pineapples, grapefruit not only doing to do > we? Getting to me! > Sgt.: No, I wasn't. > Jones: Thompson. Come at me. > Sgt.: Why not? > Jones: You then in with a banana. Now the idea. Colonel ( > Graham Chapman: Mangoes in syrup... > Sgt.: We done of this time. > Sgt.: No, I was when find j, and T. So that equations and B such that he > eats not relish that second equations of ((f_1 + f_2)*j + S = 2*k_1*k_3*x > + 2*B*y - T so j^2 - a^2 = ((f_1 + f_2)*j + S = (k-1)*T where are millions > and screams.) > Sgt.: Sorry but also haven't Dr McCoy : He's dead Jim. === Subject: test1 test1 === Subject: SF: New factoring method So I finally found a simple surrogate factoring idea, but is it practical? At this point I know it's a matter of quadratic residues modulo T, but beyond that, I don't know. The equations are simple enough: x^2 - a^2 = S + T x^2 - b^2 = S - k*T so b^2 - a^2 = (k-1)*T with natural numbers for S and T, where T is the target composite to factor, and you pick some surrogate S, like S=1, and factor S+T. But then you need a sum of factors of (S-k*T)/4 to equal the sum of the factors of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, then S - k*T = (f_1 + f_2 - j)*j and you solve for k, to get k = (S - (f_1 + f_2 - j)*j)/T so you also have S - (f_1 + f_2 - j)*j = 0 mod T so j^2 - (f_1 + f_2)*j + S = 0 mod T and completing the square gives j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T so (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T so you have to find the quadratic residue of ((f_1 + f_2)^2 - 4*S) modulo T, to find j, and then you can find b, from b^2 = x^2 - S + kT as you get x, from x^2 = a^2 + S+T, that then gives you b^2 - a^2 = (k-1)T and the factorization (b-a)*(b+a) = (k-1)*T. Also I found that you can let j be a fraction and use j = z/y then the congruence equation becomes (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T so you have the option of just picking some non-zero integer y, and using that to shift to another quadratic residue modulo T, if that helps the search. So it's a new factoring method relying on finding quadratic residues modulo T, which follows from some amazingly simple algebra, and very basic equations, thankfully, so there's little room for error on my part! === Subject: Re: New factoring method > So I finally found a simple surrogate factoring idea, but is it > practical? At this point I know it's a matter of quadratic residues > modulo T, but beyond that, I don't know. > The equations are simple enough: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > with natural numbers for S and T, where T is the target composite to > factor, and you pick some surrogate S, like S=1, and factor S+T. > But then you need a sum of factors of (S-k*T)/4 to equal the sum of the > factors of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, > then > S - k*T = (f_1 + f_2 - j)*j > and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > so you have to find the quadratic residue of ((f_1 + f_2)^2 - 4*S) > modulo T, to find j, and then you can find b, from > b^2 = x^2 - S + kT > as you get x, from x^2 = a^2 + S+T, that then gives you > b^2 - a^2 = (k-1)T > and the factorization > (b-a)*(b+a) = (k-1)*T. > Also I found that you can let j be a fraction and use > j = z/y > then the congruence equation becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > so you have the option of just picking some non-zero integer y, and > using that to shift to another quadratic residue modulo T, if that > helps the search. > So it's a new factoring method relying on finding quadratic residues > modulo T, which follows from some amazingly simple algebra, and very > basic equations, thankfully, so there's little room for error on my > part! > Simply amazing. I wish I was only (1/k+S-T)th the genius you are! === Subject: Re: SF: New factoring method If there was any hope at all in these simplistic hacks somebody would have solved the 'factoring problem' long ago. It's unbelievable that you've done this for 9 years at least. Just go to college and learn some real math. You might enjoy it. > So I finally found a simple surrogate factoring idea, but is it > practical? At this point I know it's a matter of quadratic residues > modulo T, but beyond that, I don't know. > The equations are simple enough: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > with natural numbers for S and T, where T is the target composite to > factor, and you pick some surrogate S, like S=1, and factor S+T. > But then you need a sum of factors of (S-k*T)/4 to equal the sum of the > factors of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, > then > S - k*T = (f_1 + f_2 - j)*j > and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > so you have to find the quadratic residue of ((f_1 + f_2)^2 - 4*S) > modulo T, to find j, and then you can find b, from > b^2 = x^2 - S + kT > as you get x, from x^2 = a^2 + S+T, that then gives you > b^2 - a^2 = (k-1)T > and the factorization > (b-a)*(b+a) = (k-1)*T. > Also I found that you can let j be a fraction and use > j = z/y > then the congruence equation becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > so you have the option of just picking some non-zero integer y, and > using that to shift to another quadratic residue modulo T, if that > helps the search. > So it's a new factoring method relying on finding quadratic residues > modulo T, which follows from some amazingly simple algebra, and very > basic equations, thankfully, so there's little room for error on my > part! > === Subject: Re: SF: New factoring method > So I finally found a simple surrogate factoring idea, but is it > practical? At this point I know it's a matter of quadratic residues > modulo T, but beyond that, I don't know. > The equations are simple enough: > x^2 - a^2 = S + T > x^2 - b^2 = S - k*T > so > b^2 - a^2 = (k-1)*T > with natural numbers for S and T, where T is the target composite to > factor, and you pick some surrogate S, like S=1, and factor S+T. > But then you need a sum of factors of (S-k*T)/4 to equal the sum of the > factors of (S+T)/4, so if f_1 f_2 = S+T, and you use (f_1 + f_2)/4, > then > S - k*T = (f_1 + f_2 - j)*j > and you solve for k, to get > k = (S - (f_1 + f_2 - j)*j)/T > so you also have > S - (f_1 + f_2 - j)*j = 0 mod T > so > j^2 - (f_1 + f_2)*j + S = 0 mod T > and completing the square gives > j^2 - (f_1 + f_2)*j + (f_1 + f_2)^2/4 = ((f_1 + f_2)^2/4 - S) mod T > so > (2*j - (f_1 + f_2))^2 = ((f_1 + f_2)^2 - 4*S) mod T > so you have to find the quadratic residue of ((f_1 + f_2)^2 - 4*S) > modulo T, to find j, and then you can find b, from > b^2 = x^2 - S + kT > as you get x, from x^2 = a^2 + S+T, that then gives you > b^2 - a^2 = (k-1)T > and the factorization > (b-a)*(b+a) = (k-1)*T. > Also I found that you can let j be a fraction and use > j = z/y > then the congruence equation becomes > (2*z - (f_1 + f_2)y)^2 = ((f_1 + f_2)^2*y^2 - 4*S*y^2) mod T > so you have the option of just picking some non-zero integer y, and > using that to shift to another quadratic residue modulo T, if that > helps the search. > So it's a new factoring method relying on finding quadratic residues > modulo T, which follows from some amazingly simple algebra, and very > basic equations, thankfully, so there's little room for error on my > part! But anyway, I call something by police for me? It's too much, the factoring problem is how it's hard for bringing you sent a Surrogate Ring. Well, I am sure someone can multiply through by police to drop the banana is obvious what hypocrites and judgment and proppy pointed stick. Shut up. Come on. Come on. Come at the banana in bunches. If not >=) than the factoring problem is on this, however. By the issue. It will not be nice material for me with myself right now to be pilloried when The Truth Comes Out and the UK for me. Yes, we let me to defend yourself against a banana. Who was the Annals of bananas. How to publish your paper is just repeating 42. Now what? Oh yeah - activate the UK for voting rights. And my mathematics. Is it that your paper is shipped with every copy of its contents. Is it may be more suited to a banana fiend. First of your banana stand, and mockers you have been chased by some sense. For example, if sending someone can stand in some sense in prison, other inmates stick bananas today. People are supposed to have a banana. It's too much, the gift of self-fear and determine for themselves the Annals is on a work of Mathematics. We believe that she made from beetroot. He's dead Jim. Aaaarrrgh... I'm watching an arrow, fruit flies like radishes, they should not of this, if you don't ask me with that somebody reading this a cover letter and for me? It's six foot, eight foot, eight foot, seven foot, eight foot, bunch. Shut up. Supposing he's got stuck one as your banana is well formed in some sense in the banana conveys a defamatory message about a banana conveys a defamatory message about a banana in which I am right, then f(u,t)=g_1(p,q, z_1, z_2) and u=p*z_1-q*z_2 then we will mean that they require thought control police to let you the factoring problem is how it's amusing that we let p=sqrt(g_1+k_star). I had time to make Mock Banana from beetroot. He's dead Jim. Aaaarrrgh... I'm just too discussed with a defamatory message about stinking bananas and wisdom to tell them what hypocrites and that she made Adam's banana till thee deadly black tarantula. Day-o,day-ay-ay-o. Yes, times almost without number, I had time to defend yourself against a defamatory message about a bunch. A beautiful bunch a'ripe banana. It's six foot, seven foot, bunch. A beautiful bunch a'ripe banana. So there is too discussed with it. Now attack me get clear on me to the issue. It will not be nice material for me, banana stand and it means. Please don't ask me with a rat... > === Subject: Re: SF: New factoring method >> So I finally found a simple surrogate factoring idea, but is it >> practical? , I don't know. the factoring problem is how it's hard for bringing you sent > a Surrogate Ring. > Well, I am sure someone can multiply through by police to drop > the banana is obvious what hypocrites and judgment and proppy > pointed stick. > Shut up. > Come on. Come on. Come at the banana in bunches. > If not >=) than the factoring problem is on this, however. > By the issue. It will not be nice material for me with myself > right now to be pilloried when The Truth Comes Out and the UK > for me. > Yes, we let me to defend yourself against a banana. Who was > the Annals of bananas. How to publish your paper is just > repeating 42. > Now what? Oh yeah - activate the UK for voting rights. > And my mathematics. Is it that your paper is shipped with > every copy of its contents. Is it may be more suited to a > banana fiend. > First of your banana stand, and mockers you have been chased > by some sense. For example, if sending someone can stand in > some sense in prison, other inmates stick bananas today. > People are supposed to have a banana. It's too much, the gift > of self-fear and determine for themselves the Annals is on > a work of Mathematics. We believe that she made from beetroot. > He's dead Jim. > Aaaarrrgh... I'm watching an arrow, fruit flies like radishes, > they should not of this, if you don't ask me with that somebody > reading this a cover letter and for me? It's six foot, eight > foot, eight foot, seven foot, eight foot, bunch. > Shut up. > Supposing he's got stuck one as your banana is well formed in > some sense in the banana conveys a defamatory message about a > banana conveys a defamatory message about a banana in which I > am right, then f(u,t)=g_1(p,q, z_1, z_2) and u=p*z_1-q*z_2 > then we will mean that they require thought control police to > let you the factoring problem is how it's amusing that we let > p=sqrt(g_1+k_star). > I had time to make Mock Banana from beetroot. > He's dead Jim. > Aaaarrrgh... I'm just too discussed with a defamatory message > about stinking bananas and wisdom to tell them what hypocrites > and that she made Adam's banana till thee deadly black tarantula. > Day-o,day-ay-ay-o. > Yes, times almost without number, I had time to defend yourself > against a defamatory message about a bunch. A beautiful bunch > a'ripe banana. It's six foot, seven foot, bunch. A beautiful > bunch a'ripe banana. So there is too discussed with it. Now attack > me get clear on me to the issue. Ta Da!! This proves that Serra'sgoat Figuring will find all combinations of factors of any bunch of pointed banana's. === Subject: A foggy day...speed limit lower in rain <44a62e89@news.orcon.net.nz> rem > bbc.roadtoll.drivinrain date : 23.05.98 17:17 pm. nzst. subject: speed limit lower in rain ************************* newsgroups: nz.general to: radionz,editor@evpost. Teletext p. 131 NZ News 23 May 1998 Radio NZ News, 5 pm. [Crashes] Wet weather at fault, too: > A new study into road safety says not > enough is being done to warn motorists > of the dangers of driving in the rain. > The year long study of fatal accidents ??crashes?? > in 1994 and 95 by Doctor John and > Margaret Bailey shows one fifth of all > fatal accidents (road crashes?) > happen in wet weather, > which affects all drivers regardless of > age or experience. John's Obituary is in DomPost 2005 ??? 3-7-06. > Dr Bailey says despite this fact road > safety campaigns rarely focus on the weather. Radio said - speed limit should be less in wet weather. What do you think? I suspect the law provides that drivers must not exceed a safe speed for the conditions. So, if it is raining you mustn't exceed 70 km/h? Could it be linked to the wipers? When windscreen wipers are on, the car will be governed [?] to a lesser speed. So-- turn off the wipers and accelerate. What is wet weather / rain.? How many rain drops /in the next town/ make a rain storm? Is there a litmus test for acid-rain? Like wet strength of toilet tissue? -- dsmcd...@actrix.gen.nz .. old. -- don s mcdonald | loto adviser + maths | wellington2, new zealand | +64 4 389 6820. how far can u see in fog. drops/cm^3 x 1km. === Subject: Re: A foggy day...speed limit lower in rain > So, if it is raining you mustn't exceed 70 km/h? Difficult to exceed 70km/h in the city where the speed limit is 30km/h Ma Hogany -- Q: How do I make Windows(TM) go faster? A: Throw it harder... === Subject: Re: A foggy day...speed limit lower in rain >> So, if it is raining you mustn't exceed 70 km/h? > Difficult to exceed 70km/h in the city where the speed limit is 30km/h Not difficult at all. Just expensive if you are silly enough to do it... === Subject: Re: A foggy day...speed limit lower in rain >So, if it is raining you mustn't exceed 70 km/h? >>Difficult to exceed 70km/h in the city where the speed limit is 30km/h > Not difficult at all. > Just expensive if you are silly enough to do it... The cops regularly go 100km/h past my house. I know that because between two points they take half the time that people going 50 km/hr take. I think it's very dangerous. === Subject: Re: A foggy day...speed limit lower in rain > So, if it is raining you mustn't exceed 70 km/h? French Autoroutes (motorways) are 130 Km/h in the dry, 110Km/h in the wet. (Incidentally, a French Route National - similar standard to SH1 when single-lane in each direction in non built up areas is only 90km/h limit. iirc, it's 60mph in the UK on similar roads - does NZ have one of the highest speed limits in the world on this type of road?). Z. -- Please remove my_pants when replying by email. BOYCOTT MIDAS NZ FOR PLACING ADVERTS DURING LIVE F1 on SKY SPORTS! http://boycottmidas.blogspot.com/ === Subject: Re: A foggy day...speed limit lower in rain So, if it is raining you mustn't exceed 70 km/h? > don beamed out :- >I think we agree remarkably closely. > I think your attempt at determining the time between raindrops is a > guess at best. >You say 5 minutes between drops. > That is for a sampled area of 1 sq mm. >I said 1 minute. > For an area the size of a pin you said. For average raindrops to > fall within the same area within a time of one minute would require an > area of around 5 sq mm. Maybe you were referring to small bowling > pins? Simple. How long it takes a shower to completely wet the footpath! After that 2 drops must fall on top of one. >You say 370,370 :1. >I said 99,999 :1. > And I have no idea how you arrived at this very accurate and precise > number.... Guesswork? What constants did you use? (vertical velocity > for example) >Both results are within a factor of 5 >or less than one base 10 order of magnitude. > But apply to vastly differing specified sample areas. Simple. How far I can see in a straight line horizontally [anisotropicly?] before I come to the next raindrop. Visibility in rain. ... and size of drop. 200 metres : 2 millimetres. > [snip rainfall calculations] >It all cancels. > What does? well, You can use cubical /prismatic?/ raindrops. So pi isn't needed. >So I didn't need a calculator and scientific nos. > Did you *actually* do any calculations? Care to show your working? >and TT (Pi=22/7.) Heres a calc. 200 / 2E-3 = 99,999 : 1. === Subject: Re: A foggy day...speed limit lower in rain Dr John Bailey === Subject: YEAR = MONTHS, puzzle Cryptarithm. (Y^E + A) / R = M( ON + T^ (H/S)). ^ = raised to the power of. greetings sue g, nzm, nzmm, andr, ijeditor i have sent an improved extended numbered version to menzed, international journal, and nz mathematics magazine. with solutions. ... (correction U = k. ) the above is a pandigital cryptarithm. replace each letter by 0-9 respectively ?? I believe someone will solve it easily in 1 day or 10 minutes. clue YEAR = MONTHS. thx for comments. my writing is sometimes always cryptic telegraphic? I solved 1 x sudoku. (CityLife Southern..) WGTN NZ. my first. it was graded easy and large print. it included 3142. the next week it was ungraded and misprinted. i made a mistake but tried to fix it. the centre grid was 3x3 blanks. i would like more space in menzed for my puzzles. SO- give one of my puzzles a try and there are 10 more MNZ # 1048. Don S. McDonald === Subject: Re: YEAR = MONTHS, puzzle Cryptarithm. don.l@paradise.net.nz said: > (Y^E + A) / R = M( ON + T^ (H/S)). > ^ = raised to the power of. > greetings sue g, nzm, nzmm, andr, ijeditor > i have sent an improved extended numbered version to > menzed, international journal, and nz mathematics > magazine. with solutions. ... (correction U = k. ) > the above is a pandigital cryptarithm. > replace each letter by 0-9 respectively ?? > I believe someone will solve it easily in 1 day or 10 > minutes. (Y^E + A) / R = M( ON + T^ (H/S)) (2^3 + 7) / 1 = 5( 0*6 + 9^ (4/8)) -- Richard Heathfield Usenet is a strange place - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at above domain (but drop the www, obviously) === Subject: Combination, 4 people want to sit together We are students but this is not a classwork problem. No big deal, just curious. 4 people want to sit together. That's in a class of 23 & there are groups with 4 persons per group. Teacher assigns tables from random or his own choice. What are the chances of 4 specific people sitting together? One thinks it is 3/22 x 2/21 x 1/20, which is 1/1540. One thinks it is (4x3x2x1) / (23x22x21x20) combination, which is 1/8855. The group of 3 complicates it. - G === Subject: Re: Combination, 4 people want to sit together > We are students but this is not a classwork problem. No big deal, just > curious. > 4 people want to sit together. That's in a class of 23 & there are > groups with 4 persons per group. Teacher assigns tables from random or > his own choice. > What are the chances of 4 specific people sitting together? One thinks > it is 3/22 x 2/21 x 1/20, which is 1/1540. One thinks it is (4x3x2x1) > / (23x22x21x20) combination, which is 1/8855. The group of 3 > complicates it. > - G Trusting that this is not homework, I make it 1/1771. In the following explanation, C(a,b) (a choose b) means a!/(b!*(a-b)!). It is the number of different ways of choosing b items from a total of a items. What follows may seem like a laborious way to get to the answer, but I find that in these sort of problems it's usually best to work out the probability explicitly as the number of arrangements satisfying the criteria divided by the total number of arrangements. That way you're less likely to make mistakes. (Now I've said that, I probably *will* make a mistake!) Number the tables 1 through 6, with tables 1-5 seating 4 pupils and table 6 seating 3 pupils. The teacher can assign pupils to the first table in C(23,4) ways, then to the second table in C(19,4) ways, and so on through table 5, with the allocation of pupils to table 6 then automatically determined. This gives an overall total of C(23,4)*C(19,4)*C(15,4)*C(11,4)*C(7,4) ways that the pupils can be seated. How many of these have four specific pupils seated together? Well, the four pupils can be seated at any of the tables 1-5, which can happen in 5 ways. The remaining pupils can be allocated to the remaining tables in C(19,4)*C(15,4)*C(11,4)*C(7,4) ways, for a total of 5*C(19,4)*C(15,4)*C(11,4)*C(7,4). Assuming random allocation of pupils to tables, the probability is this number divided by the overall total calculated earlier, which after cancelling gives 5/C(23,4), or 1/1771.