mm-287 Subject: Re: Simple equation system - brauche hifleI don't like porridge.-- Tony Morris(BInfTech, Cert 3 I.T., SCJP[1.4], SCJD)Software EngineerIBM Australia - Tivoli Security Software> For n in [-2.38014;9.07352] is the equation wrong...PS: I don't really like cross-postingMax>> schrieb im Newsbeitrag> Hey to all, I need to solve the following equation, but not usingtrial> and> errorany ideas?(1.2)^n > 1.6 + 0.4nI guess I fergot my basics>> Cross posting beats multiple postings all hollow.> It isn't an equation.> I reqally don't like top posting. === Subject: mathematical figuresI am a student that is wanting to type and create my own notes for class.What I would like to ask professionals out there in the physics/mathematicsfield is, what kind of software do you use to create thedrawings/illustrations that I see in math and physics text.I first draw out my illustrations and graphs using graph paper and a pencil,then try to render them in microsoft paing, so as you can imagine, mydrawings of vectographs, etc. look like crap.I have both a linux and windows machine, so any software for those platformsI can handle.ThanksDanny === Subject: Re: mathematical figures> I am a student that is wanting to type and create my own notes for class.> What I would like to ask professionals out there in the physics/mathematics> field is, what kind of software do you use to create the> drawings/illustrations that I see in math and physics text.> I first draw out my illustrations and graphs using graph paper and a pencil,> then try to render them in microsoft paing, so as you can imagine, my> drawings of vectographs, etc. look like crap.> I have both a linux and windows machine, so any software for those platforms> I can handle.> Thanks> Dannywell, I guess you could use latex for more or less every singlething you want to draw, from formulas to graph, as their is packagesdone for that purpose...but it's not the easiest way :p(however the most good looking)johann === Subject: Re: invariant, covariant, contravariant>> Co/contra variant is about how vectors and tensors transform between> coordinates.> examples?The metric tensor is covariant tensor of order two. === Subject: Re: help!!!!!Shaun:The general form of your equation is:2^(x+a)+2^(b-x)=cExpand:(2^x)(2^a)+(2^b)(1/2^x)=cDefine new terms so haveA(2^x)+B/(2^x)-c=0Multiply through by 2^x to obtain a quadratic equation in 2^x. Solve it for2^x.Bob Ferguson> hi guys,> 2^(x+1)+2^(3-x)=17> logging gives> (x+1)lg2+(3-x)lg2=lg17> expanding gives> lg2x+lg2+3lg2-lg2x=lg17> but lg2x-lg2x=0.....goinn mad...graphed it so i know x=3 but can't getthere> numerically.....please help i know i'm missing something simple> best regards> shaunSubject: Re: help!!!!! === > hi guys,> 2^(x+1)+2^(3-x)=17> logging gives> (x+1)lg2+(3-x)lg2=lg17Logging does not give this. You must log the entire side...log2( 2^(x+1)+2^(3-x) ) = log2(17)The left hand side cannot be broken into two logs.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: 128-bit GUID boundary=----=_NextPart_000_0008_01C3DEC3.33F3A3A0------------ --------------------------------------------------------- charset=iso-8859-1Hi all,I've got this math problem... I feel a little bit silly but here it is, because it seems first year CS, but my memory fails me.Sample Code:Guid myGuid = new Guid(fccf281d-47bd-45c7-8f2b-a48d462d171b);Int64 myGuidInt = BitConverter.ToInt64(myGuid.ToByteArray(),0);this.textBox1. Text = myGuid.ToString();this.textBox2.Text = myGuidInt.ToString();Byte[] me = myGuid.ToByteArray();this.textBox3.Text = ;for (int ow = 0; ow < 16; ow++){ this.textBox3.Text += me[ow].ToString() + -;}/*This generates:Int64: 5028066390298273821ByteArray: 29-40-207-252-189-71-199-69-143-43-164-141-70-45-23-27*/ Questions:The math-by-hand way of doing this is how???3. Or from the guid string fccf281d-47bd-45c7-8f2b-a48d462d171b to it's bit representation? (01010101... all 128 of them).Thank you much for any help you can give.~Js.Subject: Re: 128-bit GUID boundary=----=_NextPart_000_000F_01C3DF34.9DB38FF0 === --------- ------------------------------------------------------------ charset=iso-8859-1Guid string is in hex.You must convert each byte to hex string. For 1 byte there will be 2 hex chars.Spearators '-' are added by 8, 4, 4, 4, and 12 bytes.string retVal = ; byte[] bytes = Guid.NewGuid().ToByteArray(); foreach(byte b in bytes){ retVal += b.ToString(x); } MessageBox.Show(retVal); Hi all, I've got this math problem... I feel a little bit silly but here it is, because it seems first year CS, but my memory fails me. Sample Code: Guid myGuid = new Guid(fccf281d-47bd-45c7-8f2b-a48d462d171b); Int64 myGuidInt = BitConverter.ToInt64(myGuid.ToByteArray(),0); this.textBox1.Text = myGuid.ToString(); this.textBox2.Text = myGuidInt.ToString(); Byte[] me = myGuid.ToByteArray(); this.textBox3.Text = ; for (int ow = 0; ow < 16; ow++) { this.textBox3.Text += me[ow].ToString() + -; } /* This generates: Int64: 5028066390298273821 ByteArray: 29-40-207-252-189-71-199-69-143-43-164-141-70-45-23-27 */ Questions: The math-by-hand way of doing this is how??? 3. Or from the guid string fccf281d-47bd-45c7-8f2b-a48d462d171b to it's bit representation? (01010101... all 128 of them). Thank you much for any help you can give. ~Js. === Subject: JSH: Basic result, test of current mathematicsOne hallmark of new research is the ability to answer questions thatothers might not have even considered. I've been playing with aresult that I've mentioned before, but became curious enough to postit again, and see if I'm wrong in assuming that what's easy for me toprove, is not so easy for contemporary mathematicians usingcontemporary mathematics!Following from my research on factoring polynomials intonon-polynomial factors I have a rather simple result that given primeintegef_1, f_2, and integer M, whereM = f_1 f_2if you find integers x and y, wherex^j = -y^j mod Mwhere j is a positive odd integer,then it must be true that (x+y) = 0 mod f_1 or(x+y) = 0 mod f_2.I'd like to see someone prove that result, if you can.I consider this a rather easy test. === Subject: Re: JSH: Basic result, test of current mathematics> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!Well it's always embarrassing to make such grand statements and laterfind out that you don't have something spectacular, and even worsethat you've made a mistake.I made a mistake with something I thought I'd proven, but clearly hadnot as revealed by some counterxamples pos in response to thethreads I crea.So what could motivate a person to put something out there and see ifit flies?I suggest to you that it's curiosity and excitement. It's simply areality that I've learned to accept that I get exci by these littleadventures in mathematics, trying to find something rather simple thatis extraordinary.What I've often found in response is a hate reaction, which I findtroubling.Sure, I can understand annoyance, but hey, it's Usenet!!!What motivates certain posters to obessively reply to my posts with aclear intent to cause me emotional harm?I think it's them, not me.I've read posts at times where people chortle about me being an *easy*target because I'm so often wrong. It's like they salivate at theidea that more than likely something I'll say, or much of what I saywill be wrong so they can attack!It's as if there are people here on Usenet who live for the thrill oftrying to hurt other people who have it in their minds that I'm a niceperson to kick because I clearly enjoy chasing after hard mathematics,and often make mistakes.Such people have left a clear trail on Usenet, and I suggest to otherreaders that it's not really fair to attack someone, who comes on a*public* forum, makes posts, even if wacky or wrong, showing anexcitement about the discipline without admitting that you're sayingsomething about yourself.Sure others have often complained that I've also shown a churlishside, hurling insults myself, but the beauty of Usenet is the abilityto ignore posters.It's like--turn the channel.The posters who have demonstra a glee at trying to inflictemotional pain on me have demonstra their own monstrosity, andgiven you some sense of what they're capable of doing.I suggest to you that they have demonstra that if possible, ifsociety allowed, they might engage in even more direct hostilebehavior. Who knows what some of these people have done that's beenhidden. Who knows where they learned their lessons of hatred.Some people just like to try and hurt othewhile most do their bestto get along. It seems to me that what I've done is get exci aboutsome ideas, many of them wrong, and unfortunately attract a hateminded crowd of people not afraid to let you all know that yes, theylike to try to hurt otheand trying to get along, is less importantthan their real agenda.To them Usenet is a hunting ground, and other people their prey.I've decided not to let them bug me, and to do my best to ignore them,and enjoy posting. Hey, I'm sure I'll keep getting things wrong, butit's Usenet.If you don't like it, turn the channel. === Subject: Re: JSH: Basic result, test of current mathematics> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> Well it's always embarrassing to make such grand statements and later> find out that you don't have something spectacular, and even worse> that you've made a mistake.> I made a mistake with something I thought I'd proven, but clearly had> not as revealed by some counterxamples pos in response to the> threads I crea.> So what could motivate a person to put something out there and see if> it flies?> I suggest to you that it's curiosity and excitement. It's simply a> reality that I've learned to accept that I get exci by these little> adventures in mathematics, trying to find something rather simple that> is extraordinary.> What I've often found in response is a hate reaction, which I find> troubling.> Sure, I can understand annoyance, but hey, it's Usenet!!!> What motivates certain posters to obessively reply to my posts with a> clear intent to cause me emotional harm?> I think it's them, not me.> I've read posts at times where people chortle about me being an *easy*> target because I'm so often wrong. It's like they salivate at the> idea that more than likely something I'll say, or much of what I say> will be wrong so they can attack!> It's as if there are people here on Usenet who live for the thrill of> trying to hurt other people who have it in their minds that I'm a nice> person to kick because I clearly enjoy chasing after hard mathematics,> and often make mistakes.> Such people have left a clear trail on Usenet, and I suggest to other> readers that it's not really fair to attack someone, who comes on a> *public* forum, makes posts, even if wacky or wrong, showing an> excitement about the discipline without admitting that you're saying> something about yourself.> Sure others have often complained that I've also shown a churlish> side, hurling insults myself, but the beauty of Usenet is the ability> to ignore posters.> It's like--turn the channel.> The posters who have demonstra a glee at trying to inflict> emotional pain on me have demonstra their own monstrosity, and> given you some sense of what they're capable of doing.> I suggest to you that they have demonstra that if possible, if> society allowed, they might engage in even more direct hostile> behavior. Like e-mailing someone's employer false racism allegations?> Who knows what some of these people have done that's been> hidden. Who knows where they learned their lessons of hatred.> Some people just like to try and hurt othewhile most do their best> to get along. It seems to me that what I've done is get exci about> some ideas, many of them wrong, and unfortunately attract a hate> minded crowd of people not afraid to let you all know that yes, they> like to try to hurt otheand trying to get along, is less important> than their real agenda.> To them Usenet is a hunting ground, and other people their prey.> I've decided not to let them bug me, and to do my best to ignore them,> and enjoy posting. Hey, I'm sure I'll keep getting things wrong, but> it's Usenet.> If you don't like it, turn the channel.> === Subject: Re: JSH: Basic result, test of current mathematics[snip]> Well it's always embarrassing to make such grand statements and later> find out that you don't have something spectacular, and even worse> that you've made a mistake.Oh, is it? Then why persist in the embarrassing behavior?> I made a mistake with something I thought I'd proven, but clearly had> not as revealed by some counterxamples pos in response to the> threads I crea.> So what could motivate a person to put something out there and see if> it flies?If it flies? Or if it draws flies?> I suggest to you that it's curiosity and excitement. It's simply a> reality that I've learned to accept that I get exci by these little> adventures in mathematics, trying to find something rather simple that> is extraordinary.I suggest it is the behavior of an arrogant blowhard who believes his flashes of brilliance are beyondchallenge and only looks closely at his own mistakes when someone succeeds in rubbing his nose in them.If you get 'exci' about posting false claims which you insist you can prove, you need 24 hoursupervision.> What I've often found in response is a hate reaction, which I find> troubling.Why do you find it troubling? Those responses have been to provide counterexamples to your false claims.You SHOULD welcome them. Any hostility in the responses has been due to problems with your behavior, notthe errors in your math.> Sure, I can understand annoyance, but hey, it's Usenet!!!> What motivates certain posters to obessively reply to my posts with a> clear intent to cause me emotional harm?If pointing out your errors causes you emotional harm try another newsgroup.> I think it's them, not me.Of course you don't think it's you. You don't think!> I've read posts at times where people chortle about me being an *easy*> target because I'm so often wrong. It's like they salivate at the> idea that more than likely something I'll say, or much of what I say> will be wrong so they can attack!No. Of course it's true that most of what you say will be wrong. But the key point is that most of what yousay, even if it's wrong, you will defend vehemently, militantly and passionately -- even if it leads toyour cursing your critics and hurling gutter-language invectives or threats at them.> It's as if there are people here on Usenet who live for the thrill of> trying to hurt other people who have it in their minds that I'm a nice> person to kick because I clearly enjoy chasing after hard mathematics,> and often make mistakes.That's NOT all you do. You also promote your own unsuppor excellence with suggestions that your are somekind of superhuman suffering from outrageous attacks by subhuman conspirators. You are completelydelusional.> Such people have left a clear trail on Usenet, and I suggest to other> readers that it's not really fair to attack someone, who comes on a> *public* forum, makes posts, even if wacky or wrong, showing an> excitement about the discipline without admitting that you're saying> something about yourself.You don't come to the forum with just 'wacky or wrong' posts. You also come with an armory of insults andthreats, ready for launch at anyone who points out your errors. Who do you think is going to sign up forthis new *humble* attitude of yours?> Sure others have often complained that I've also shown a churlish> side, hurling insults myself, but the beauty of Usenet is the ability> to ignore posters.Then ignore them. Your post here is largely concerned with your complaints about posters you don't like.Follow your own advice, hypocrite!> It's like--turn the channel.Then turn it.> The posters who have demonstra a glee at trying to inflict> emotional pain on me have demonstra their own monstrosity, and> given you some sense of what they're capable of doing.It's more like poking a dead dog with a stick -- mostly curiosity.> I suggest to you that they have demonstra that if possible, if> society allowed, they might engage in even more direct hostile> behavior. Who knows what some of these people have done that's been> hidden. Who knows where they learned their lessons of hatred.Whoa! Back up the truck, . *You* are the one who has made repea threats to other posters.These are on public record and can be recovered at a moments notice. What evidence of 'direct hostilebehavior' have you seen from other posters?> Some people just like to try and hurt othewhile most do their best> to get along. It seems to me that what I've done is get exci about> some ideas, many of them wrong, and unfortunately attract a hate> minded crowd of people not afraid to let you all know that yes, they> like to try to hurt otheand trying to get along, is less important> than their real agenda.Aw, shucks, folks. If you put it that way, you're not such a bad guy after all. You and Rodney King seem tohave the right idea. By the way, what is their *real* agenda?> To them Usenet is a hunting ground, and other people their prey.Oh, so that's it. But to most readethe 'sci.math' forum is a technical forum, not a sports arena. Mostof the psychobabble has come from you.> I've decided not to let them bug me, and to do my best to ignore them,> and enjoy posting. Hey, I'm sure I'll keep getting things wrong, but> it's Usenet.> If you don't like it, turn the channel.> James it's-hard-to-be-humble HarrisWacky, isn't it? But, hey, it's just mathematics. Yup, yup, yup.--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: Re: JSH: Basic result, test of current mathematics>> One hallmark of new research is the ability to answer questions that>> others might not have even considered. I've been playing with a>> result that I've mentioned before, but became curious enough to post>> it again, and see if I'm wrong in assuming that what's easy for me to>> prove, is not so easy for contemporary mathematicians using>> contemporary mathematics!>Well it's always embarrassing to make such grand statements and later>find out that you don't have something spectacular, and even worse>that you've made a mistake.If you find it embarrasing then why do you keep _doing_ it?Over and over and over and over...>I made a mistake with something I thought I'd proven, but clearly had>not as revealed by some counterxamples pos in response to the>threads I crea.>So what could motivate a person to put something out there and see if>it flies?>I suggest to you that it's curiosity and excitement. It's simply a>reality that I've learned to accept that I get exci by these little>adventures in mathematics, trying to find something rather simple that>is extraordinary.Suggest what you want. The actual explanation is clear to everyoneelse, _regardless_ of what you suggest.>What I've often found in response is a hate reaction, which I find>troubling.If you'd simply posting the erroneous result people wouldsimply have pos the counterexample. The hate reactioncomes from your posting the result _together_ with thosecomments about how it will be interesting to see whetheranyone's smart enough to be able to prove this thingthat you proved.>Sure, I can understand annoyance, but hey, it's Usenet!!!>What motivates certain posters to obessively reply to my posts with a>clear intent to cause me emotional harm?>I think it's them, not me.>[...]>To them Usenet is a hunting ground, and other people their prey.>I've decided not to let them bug me, and to do my best to ignore them,>and enjoy posting. Every once in a while you say something like this.Then a little later you come back with something likeYou stupid head!!! What the is wrong with you Ullrich?No matter how many ing times I tell you to off, you keepreplying to me!!!What the is your problem you head?You Ullrich are a stupid piece of dumb who refuses to get themessage when someone does NOT want to talk to you, you stupid ingty asshole.You are an ASSHOLE Ullrich!!! Now why don't you take your dumb assstupid self somewhere to GET A ING CLUE and QUIT ING REPLYINGTO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!! OFF!!!!Can't you get it through your stupid head? OFF!!!!!!!!!!!!!!!!!And then every once in a while you post something wonderingwhy nobody wuves you, poor baby.> Hey, I'm sure I'll keep getting things wrong, but>it's Usenet.>If you don't like it, turn the channel.>************************ === Subject: Re: JSH: Basic result, test of current mathematics> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer,> then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2.> I'd like to see someone prove that result, if you can.> I consider this a rather easy test.> It is remarkable how James expects others to do all his homework for him after the way he treats everyone. === Subject: Re: JSH: Basic result, test of current mathematics>It is remarkable how James expects others to do all his homework for him >after the way he treats everyone.And still more remarkable that he seems to be quite correct in thatexpectation.-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules! === Subject: Re: JSH: Basic result, test of current mathematics>One hallmark of new research is the ability to answer questions that>others might not have even considered. I've been playing with a>result that I've mentioned before, but became curious enough to post>it again, and see if I'm wrong in assuming that what's easy for me to>prove, is not so easy for contemporary mathematicians using>contemporary mathematics!Funny you should mention that. I have several elementary proofs that 1 and2 are, in fact, equal, and none of these proofs are possible (let alone easy) with contemporary mathematics.Perhaps a symposium is necessary. === Subject: Re: JSH: Basic result, test of current mathematics>One hallmark of new research is the ability to answer questions that>others might not have even considered. I've been playing with a>result that I've mentioned before, but became curious enough to post>it again, and see if I'm wrong in assuming that what's easy for me to>prove, is not so easy for contemporary mathematicians using>contemporary mathematics!> Funny you should mention that. I have several elementary proofs that 1 and> 2 are, in fact, equal, and none of these proofs are possible (let alone > easy) with contemporary mathematics.> Perhaps a symposium is necessary.Although this is merely more tiresome Harris-bashing, it does raisean interesting possibility. I have poin out to James the curiouscoincidence that the 2005 Joint Mathematics Meeting is in his homestate of Georgia, suggesting that he might enjoy presenting his workthere. What hadn't occurred to me is that he might instead try toorganize a mini-symposium on the fundamental defect in the algebraicintegeor on some other topic of interest to him.Oh sure, he's probably worried that the mathematical establishmentwouldn't allow him to do it, but life rewards those who try. AsJames well knows. === Subject: Re: JSH: Basic result, test of current mathematics Adjunct Assistant Professor at the University of Montana.>>One hallmark of new research is the ability to answer questions that>>others might not have even considered. I've been playing with a>>result that I've mentioned before, but became curious enough to post>>it again, and see if I'm wrong in assuming that what's easy for me to>>prove, is not so easy for contemporary mathematicians using>>contemporary mathematics!>> Funny you should mention that. I have several elementary proofs that 1 and>> 2 are, in fact, equal, and none of these proofs are possible (let alone >> easy) with contemporary mathematics.>> Perhaps a symposium is necessary.>Although this is merely more tiresome Harris-bashing, it does raise>an interesting possibility. Of course, it is possible that is referring to the technicalmeaning of symposium (hence the use of that rather thanconference): symposium comes from the Greek symposion/sympinein, todrink together. Hence a 'convival party', like the parties afterbanquets in Greece...-- === Subject: Re: JSH: Basic result, test of current mathematics|One hallmark of new research is the ability to answer questions that|others might not have even considered. I've been playing with a|result that I've mentioned before, but became curious enough to post|it again, and see if I'm wrong in assuming that what's easy for me to|prove, is not so easy for contemporary mathematicians using|contemporary mathematics!||Following from my research on factoring polynomials into|non-polynomial factors I have a rather simple result that given prime|integef_1, f_2, and integer M, where||M = f_1 f_2||if you find integers x and y, where||x^j = -y^j mod M||where j is a positive odd integer,||then it must be true that ||(x+y) = 0 mod f_1 or||(x+y) = 0 mod f_2.||I'd like to see someone prove that result, if you can.||I consider this a rather easy test.f1=7,f2=13,j=3,x=1,y=75-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: JSH: Basic result, test of current mathematics> |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given prime> |integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> f1=7,f2=13,j=3,x=1,y=75Good counterexample. I thought I had a fix, but realize now that I'm wrong here.Thanks!Subject: Re: JSH: Basic result, test of current mathematics === > |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given prime> |integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> f1=7,f2=13,j=3,x=1,y=75Nice counter-example. I'm curious to see how James genera his result. With these numbeperhaps we can help him see where he made his mistake.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: JSH: Basic result, test of current mathematics>> |One hallmark of new research is the ability to answer questions that>> |others might not have even considered. I've been playing with a>> |result that I've mentioned before, but became curious enough to post>> |it again, and see if I'm wrong in assuming that what's easy for me to>> |prove, is not so easy for contemporary mathematicians using>> |contemporary mathematics!>> |>> |Following from my research on factoring polynomials into>> |non-polynomial factors I have a rather simple result that given prime>> |integef_1, f_2, and integer M, where>> |>> |M = f_1 f_2>> |>> |if you find integers x and y, where>> |>> |x^j = -y^j mod M>> |>> |where j is a positive odd integer,>> |>> |then it must be true that >> |>> |(x+y) = 0 mod f_1 or>> |>> |(x+y) = 0 mod f_2.>> |>> |I'd like to see someone prove that result, if you can.>> |>> |I consider this a rather easy test.>> f1=7,f2=13,j=3,x=1,y=75>Nice counter-example. I'm curious to see how James genera his >result. With these numbeperhaps we can help him see where he made >his mistake.Guffaw. Help him see where he made his mistake? You simplyhaven't been paying attention.************************ === Subject: Re: JSH: Basic result, test of current mathematics Adjunct Assistant Professor at the University of Montana. [.snip.]>> |Following from my research on factoring polynomials into>> |non-polynomial factors I have a rather simple result that given prime>> |integef_1, f_2, and integer M, where>> |>> |M = f_1 f_2>> |>> |if you find integers x and y, where>> |>> |x^j = -y^j mod M>> |>> |where j is a positive odd integer,>> |>> |then it must be true that >> |>> |(x+y) = 0 mod f_1 or>> |>> |(x+y) = 0 mod f_2.>> |>> |I'd like to see someone prove that result, if you can.>> |>> |I consider this a rather easy test.>> f1=7,f2=13,j=3,x=1,y=75>Nice counter-example. I'm curious to see how James genera his >result. With these numbeperhaps we can help him see where he made >his mistake.It is of course dangerous to guess, but here goes.If x^j = -y^j (mod M), then x^j + y^j = 0 (mod M). We can factor x^j +y^j asx^j + y^j = (x+y)(x^{j-1} - x^{j-2}y + x^{j-3}y^2 + .... +y^{j-1}).The product is 0 (mod M), therefore is 0 modulo f_1 and modulo f_2. Aproduct is zero modulo a prime if and only if at least one factor is 0modulo a prime, so we conclude that(x+y) = 0 (mod f_1); or(x+y) = 0 (mod f_2); or(x^{j-1} - x^{j-2}y + ... + y^{j-1}) = 0 (mod M)(the last congruence because if neither x+y = 0 (mod f_1) nor x+y=0(mod f_2), then the second factor is 0 modulo both f_1 and f_2, hencemodulo lcm(f_1,f_2)=f_1*f_2=M). And for some reason the lastcongruence is thought to be impossible, thus concluding the first two.-- === Subject: Re: JSH: Basic result, test of current mathematics Adjunct Assistant Professor at the University of Montana.>> |One hallmark of new research is the ability to answer questions that>> |others might not have even considered. I've been playing with a>> |result that I've mentioned before, but became curious enough to post>> |it again, and see if I'm wrong in assuming that what's easy for me to>> |prove, is not so easy for contemporary mathematicians using>> |contemporary mathematics!>> |>> |Following from my research on factoring polynomials into>> |non-polynomial factors I have a rather simple result that given prime>> |integef_1, f_2, and integer M, where>> |>> |M = f_1 f_2>> |>> |if you find integers x and y, where>> |>> |x^j = -y^j mod M>> |>> |where j is a positive odd integer,>> |>> |then it must be true that >> |>> |(x+y) = 0 mod f_1 or>> |>> |(x+y) = 0 mod f_2.>> |>> |I'd like to see someone prove that result, if you can.>> |>> |I consider this a rather easy test.>> f1=7,f2=13,j=3,x=1,y=75>Nice counter-example. I'm curious to see how James genera his >result. With these numbeperhaps we can help him see where he made >his mistake.Unlikely; his correction amoun to adding the condition abs(x)>1,abs(y)>1, followed by an assertion of having a proof.But a quick twiddling of James Dolan result also shows that conditionto be insufficient: f1=7, f2-13, j-3, x=92, y=75.-- === Subject: Re: JSH: Basic result, test of current mathematics: Following from my research on factoring polynomials intoWhich, as you poin out yourself, was incorrect...: non-polynomial factors I have a rather simple result that given: prime integef_1, f_2, and integer M, whereYou have a result? Does this mean you proved it or you conjecture it?: I'd like to see someone prove that result, if you can.I'm sure you would. Perhaps you should phrase it Does anyone know ifthis is true, and if so, is there a simple proof? That might be better.: I consider this a rather easy test.Of what?Justin === Subject: JSH: Basic result, test REVISEDWouldn't you know I pos and someone caught me on something rightoff, so I'm revising the test, as in fact there was something I hadn'trealized!!!Thanks to James Dolan for his example with f_1 = 7, f_2 = 13, x=1,y=75.One hallmark of new research is the ability to answer questions thatothers might not have even considered. I've been playing with aresult that I've mentioned before, but became curious enough to postit again, and see if I'm wrong in assuming that what's easy for me toprove, is not so easy for contemporary mathematicians usingcontemporary mathematics!Following from my research on factoring polynomials intonon-polynomial factors I have a rather simple result that given primeintegef_1, f_2, and integer M, whereM = f_1 f_2if you find integers x and y, wherex^j = -y^j mod Mwhere j is a positive odd integer, and abs(x)>1 and abs(y)>1,then it must be true that (x+y) = 0 mod f_1 or(x+y) = 0 mod f_2.I'd like to see someone prove that result, if you can.I consider this a rather easy test.And to answer another poster who replied in my previous thread, yes, Ihave the proof.The question here is, do any of you? === Subject: Re: JSH: Basic result, test REVISED> Wouldn't you know I pos and someone caught me on something right> off, so I'm revising the test, as in fact there was something I hadn't> realized!!!> Thanks to James Dolan for his example with f_1 = 7, f_2 = 13, x=1,> y=75.> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, and abs(x)>1 and abs(y)>1,> then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2.> I'd like to see someone prove that result, if you can. I admit it. I can't do it.> I consider this a rather easy test. Let's see ... how about f_1 = 31, f_2 = 71, j = 5,x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ? Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ?Or x = 91, y = 74 ? Or x = 95, y = 72 ? You're right. It's an easy test. > And to answer another poster who replied in my previous thread, yes, I> have the proof. It's an easy that you failed. But feel free to expound at length about your proof.> The question here is, do any of you? See above. > === Subject: Re: JSH: Basic result, test REVISED>Wouldn't you know I pos and someone caught me on something right>off, so I'm revising the test, as in fact there was something I hadn't>realized!!!How is that possible? I mean you _proved_ the result, right?>Thanks to James Dolan for his example with f_1 = 7, f_2 = 13, x=1,>y=75.>One hallmark of new research is the ability to answer questions that>others might not have even considered. I've been playing with a>result that I've mentioned before, but became curious enough to post>it again, and see if I'm wrong in assuming that what's easy for me to>prove, is not so easy for contemporary mathematicians using>contemporary mathematics!Jesus. You post this pompous condescending bull about howyou're curious whether anyone is as smart as you. It takes lessthan 20 minutes for someone to find an example that showsthat you're simply _wrong_. Now when you post a correctionit's _supposed_ to be with an appropriate does of humility,but instead, even though you've just been shown wrong,you _repeat_ the condescending bull about how you'recurious whether anyone is as smart as you.And then it turns out the correc version is wrongagain.You're really never going to catch on to this. Youevidently simply don't understand what a proof is.We all make mistakes, but we don't prove thingsthat turn out to be false over and over and overand over.>Following from my research on factoring polynomials into>non-polynomial factors I have a rather simple result that given prime>integef_1, f_2, and integer M, where>M = f_1 f_2>if you find integers x and y, where>x^j = -y^j mod M>where j is a positive odd integer, and abs(x)>1 and abs(y)>1,>then it must be true that >(x+y) = 0 mod f_1 or>(x+y) = 0 mod f_2.>I'd like to see someone prove that result, if you can.>I consider this a rather easy test.>And to answer another poster who replied in my previous thread, yes, I>have the proof.No, you don't.>The question here is, do any of you?>************************ === Subject: Re: JSH: Basic result, test REVISED>And to answer another poster who replied in my previous thread, yes, I>have the proof.You had the proof last time, too. Remember? You told us so. === Subject: Re: JSH: Basic result, test REVISED>Wouldn't you know I pos and someone caught me on something right>off, so I'm revising the test, as in fact there was something I hadn't>realized!!!Yeah, go figure. === Subject: Re: JSH: Basic result, test REVISED|One hallmark of new research is the ability to answer questions that|others might not have even considered. I've been playing with a|result that I've mentioned before, but became curious enough to post|it again, and see if I'm wrong in assuming that what's easy for me to|prove, is not so easy for contemporary mathematicians using|contemporary mathematics!||Following from my research on factoring polynomials into|non-polynomial factors I have a rather simple result that given prime|integef_1, f_2, and integer M, where||M = f_1 f_2||if you find integers x and y, where||x^j = -y^j mod M||where j is a positive odd integer, and abs(x)>1 and abs(y)>1,||then it must be true that ||(x+y) = 0 mod f_1 or||(x+y) = 0 mod f_2.||I'd like to see someone prove that result, if you can.||I consider this a rather easy test.||And to answer another poster who replied in my previous thread, yes, I|have the proof.||The question here is, do any of you?f1=7,f2=13,j=3,x=53,y=75-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: JSH: Basic result, test REVISED> |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given prime> |integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer, and abs(x)>1 and abs(y)>1,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> |> |And to answer another poster who replied in my previous thread, yes, I> |have the proof.> |> |The question here is, do any of you?> f1=7,f2=13,j=3,x=53,y=75Well I'm impressed you found that counterexample in 76 minutes! Could you explain how you did it to those of us who don't know where to start?Mark Atherton === Subject: Re: JSH: Basic result, test REVISED|> f1=7,f2=13,j=3,x=53,y=75|> ||Well I'm impressed you found that counterexample in 76 minutes! Could |you explain how you did it to those of us who don't know where to start?the chinese remainder theorem makes it easy to answer lots ofquestions about arithmetic modulo a product of pairwise relativelyprime factors. i did use a calculator to check my calculations butmostly i did them in my head and it only took a few minutes to do themand anyone who used the same trick would probably find a solutionequally fast.the other trick i used was that after deciding to try j=3 it seemed tomake sense to focus on primes of the form 6n+1 because cubing modulo aprime of the form 6n+1 can be non-invertible. or something like that.-- [e-mail address jdolan@math.ucr.edu] === Subject: Re: JSH: Basic result, test REVISED> |> f1=7,f2=13,j=3,x=53,y=75> ||> |Well I'm impressed you found that counterexample in 76 minutes! Could > |you explain how you did it to those of us who don't know where to start?> the chinese remainder theorem makes it easy to answer lots of> questions about arithmetic modulo a product of pairwise relatively> prime factors. i did use a calculator to check my calculations but> mostly i did them in my head and it only took a few minutes to do them> and anyone who used the same trick would probably find a solution> equally fast.> the other trick i used was that after deciding to try j=3 it seemed to> make sense to focus on primes of the form 6n+1 because cubing modulo a> prime of the form 6n+1 can be non-invertible. or something like that.Basically he claims that if x^3 + y^3 is divisible by ab for two primes ab, then x+y is divisible by a or b. x^3 + y^3 = (x+y) (x^2 - xy + y^2), so we would find a counterexample if (x^2 - xy + y^2) is divisible by ab but x+y isn't divisible by either a or b. In other words, for a counterexample all you need is find x and y such that (x^2 - xy + y^2) has at least two prime factors that are not prime factors of x+y. If you substitute z = x+y, z^2 = x^2 + y^2 + 2xy, then (x^2 - xy + y^2) = z^2 - 3xy = z^2 - 3(z-y)y, so you need to find z,y such that z^2 - 3(z-y)y has at least two prime factors that are not prime factors of z. There are plenty of examples of that; the first are z=11 and y=1 or 3, z=13 and y=1 or 5 etc. I used a spreadsheet to find them There are quite a few cases that are no counterexamples; for example it is easy to show that if z^2 - 3(z-y)y is divisible by 2, 3 or 5 then so is z, so z^2 - 3(z-y)y must have at least two prime factors other than 2, 3 or 5. is now making wild guesses what extra conditions could exclude the counterexamples. His latest guess is that ab must not be the sum or difference of two cubes; most likely because this excludes the case ab = 7*13 which gets rid of a few counterexamples. No wonder that his would guesses don't get him anywhere. === Subject: Re: JSH: Basic result, test REVISED>Well I'm impressed you found that counterexample in 76 minutes! Could >you explain how you did it to those of us who don't know where to start?Even without any understanding of the problem, counter-examples areeasily found with a program. Something like this:#define f_max 30#define xy_max 100#define j_max 5int primes[] = {2,3,5,7,11,13,17,19,23,29,31};int power_mod(int b, int e, int m){ int i, r = b % m; for(i=1; i |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given prime> |integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer, and abs(x)>1 and abs(y)>1,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> |> |And to answer another poster who replied in my previous thread, yes, I> |have the proof.> |> |The question here is, do any of you?> f1=7,f2=13,j=3,x=53,y=75> Well I'm impressed you found that counterexample in 76 minutes! Could > you explain how you did it to those of us who don't know where to start?You write a C program. This is just being typed in, without testing. There may be a bug or two, and it only tries the case j=3.#include // This function takes a number M, returns true if M is the product // of exactly two primes which will be stored in *f1 and *f2; returns// false if M is not the product of exactly two primes. int two_factors (int M, int* f1, int* f2) { int i, j, k; for (i = 2; i*i <= M; ++i) { if (M % i != 0) continue; k = M / i; for (j = i;j*j <=k; ++j) { if (k % j == 0) break; } if (j*j > k) { *f1 = i; *f2 = j; return 1; } return 0;}int main (void) { int M; for (M = 4; M <= 10000; ++M) { int f1, f2; if (! two_factors (M, &f1, &f2)) continue; int x, y; for (x = 2; x <= M-2; ++x) { for (y = 2; y <= M-2; ++y) { if ((x*x*x + y*y*y) % M == 0) { if (x+y % f1 != 0 && x+y % f2 != 0) { printf (f1=%d, f2=%d, j=3, x=%d, y=%dn, f1, f2, x, y); } } } } } return 0;} === Subject: Re: JSH: Basic result, test REVISED> |One hallmark of new research is the ability to answer questions that> |others might not have even considered. I've been playing with a> |result that I've mentioned before, but became curious enough to post> |it again, and see if I'm wrong in assuming that what's easy for me to> |prove, is not so easy for contemporary mathematicians using> |contemporary mathematics!> |> |Following from my research on factoring polynomials into> |non-polynomial factors I have a rather simple result that given prime> |integef_1, f_2, and integer M, where> |> |M = f_1 f_2> |> |if you find integers x and y, where> |> |x^j = -y^j mod M> |> |where j is a positive odd integer, and abs(x)>1 and abs(y)>1,> |> |then it must be true that > |> |(x+y) = 0 mod f_1 or> |> |(x+y) = 0 mod f_2.> |> |I'd like to see someone prove that result, if you can.> |> |I consider this a rather easy test.> |> |And to answer another poster who replied in my previous thread, yes, I> |have the proof.> |> |The question here is, do any of you?> f1=7,f2=13,j=3,x=53,y=75Yeah, there's just no way around it. I thought I had a proof of thatresult, but now it's more interesting.Thanks!!! === Subject: Re: JSH: Basic result, test REVISED>Yeah, there's just no way around it. I thought I had a proof of that>result, but now it's more interesting.Actually, it's trivially just as interesting. === Subject: Re: JSH: Basic result, test REVISED> Wouldn't you know I pos and someone caught me on something right> off, so I'm revising the test, as in fact there was something I hadn't> realized!!!Are you the ONLY one surprised??? (Bump the Oops! counter once more!)--What a maroon! -- Bugs Bunny.----http://www.crbond.com === Subject: JSH: Basic result, test REVISION 2Wouldn't you know, I decided that it was wrong and saw what seems tobe an important condition I'd missed.Thanks to James Dolan for his examples withM=91, as apparently anexception occurs with j=3, if there exists integers c and d, such thatM - c^3 = d^3.With 91, you have c=4, d=3.One hallmark of new research is the ability to answer questions thatothers might not have even considered. I've been playing with aresult that I've mentioned before, but became curious enough to postit again, and see if I'm wrong in assuming that what's easy for me toprove, is not so easy for contemporary mathematicians usingcontemporary mathematics!Following from my research on factoring polynomials intonon-polynomial factors I have a rather simple result that given primeintegef_1, f_2, and integer M, whereM = f_1 f_2if you find integers x and y, wherex^j = -y^j mod Mwhere j is a positive odd integer, then it must be true that (x+y) = 0 mod f_1 or(x+y) = 0 mod f_2,if there doesn't exist integers c and d, such that M - c^3 = d^3.I'd like to see someone prove that result, if you can.I considered this to be a rather easy test, until I had problemsgiving it! === Subject: Re: JSH: Basic result, test REVISION 2> Wouldn't you know, I decided that it was wrong and saw what seems to> be an important condition I'd missed.> Thanks to James Dolan for his examples withM=91, as apparently an> exception occurs with j=3, if there exists integers c and d, such that> M - c^3 = d^3.> With 91, you have c=4, d=3.> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2,> if there doesn't exist integers c and d, such that M - c^3 = d^3.> I'd like to see someone prove that result, if you can.Just post what you have found so far, and we will tell you where you are wrong. === Subject: Re: JSH: Basic result, test REVISION 2 > Thanks to James Dolan for his examples withM=91, as apparently an > exception occurs with j=3, if there exists integers c and d, such that > M - c^3 = d^3.You may also thank Nora Baron. > M = f_1 f_2 > if you find integers x and y, where > x^j = -y^j mod M > where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or > (x+y) = 0 mod f_2, > if there doesn't exist integers c and d, such that M - c^3 = d^3. > Let's see ... how about f_1 = 31, f_2 = 71, j = 5, > x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ? > Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ? > Or x = 91, y = 74 ? Or x = 95, y = 72 ?So, now what? Perhaps you meant: > M = f_1 f_2 > if you find integers x and y, where > x^j = -y^j mod M > where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or > (x+y) = 0 mod f_2, > if there doesn't exist integers c and d, such that M - c^j = d^j.But that also appears to be false. (As far as I can see neither 31*71nor 29*43 can be written as sum or difference of two cubes, the firstcan not be written as sum or difference of fifth powers and the secondcan not be written as sum or difference of seventh powers.)Where is your proof?-- === Subject: Re: JSH: Basic result, test REVISION 2> Thanks to James Dolan for his examples withM=91, as apparently an> exception occurs with j=3, if there exists integers c and d, such that> M - c^3 = d^3.> You may also thank Nora Baron. Not likely! > M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2,> if there doesn't exist integers c and d, such that M - c^3 = d^3.> Let's see ... how about f_1 = 31, f_2 = 71, j = 5,> x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ?> Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ?> Or x = 91, y = 74 ? Or x = 95, y = 72 ?> So, now what? Perhaps you meant:> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2,> if there doesn't exist integers c and d, such that M - c^j = d^j.> But that also appears to be false. (As far as I can see neither 31*71> nor 29*43 can be written as sum or difference of two cubes, the first> can not be written as sum or difference of fifth powers and the second> can not be written as sum or difference of seventh powers.)> Where is your proof? There may actually be an interesting theorem in here somewhere.I found that, for j = 3, I could find lots of counterexampleswhen f_1 and f_2 were primes congruent to 1 mod 6, but I couldn'tfind any when either f_1 or f_2 or both were congruent to 5 mod 6.Similarly, when j = 5, I could find counterexamples when f_1 and f_2 were congruent to 1 mod 10, but not otherwise. Notthat I did really extensive searches. I don't see an immediate proof. === Subject: Re: JSH: Basic result, test REVISION 2>> Thanks to James Dolan for his examples withM=91, as apparently an>> exception occurs with j=3, if there exists integers c and d, such that>> M - c^3 = d^3.>>You may also thank Nora Baron.> Not likely!>> M = f_1 f_2>> if you find integers x and y, where>> x^j = -y^j mod M>> where j is a positive odd integer, >> then it must be true that >> (x+y) = 0 mod f_1 or>> (x+y) = 0 mod f_2,>> if there doesn't exist integers c and d, such that M - c^3 = d^3.Let's see ... how about f_1 = 31, f_2 = 71, j = 5,>> x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ?>> Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ?>> Or x = 91, y = 74 ? Or x = 95, y = 72 ?>>So, now what? Perhaps you meant:M = f_1 f_2>> if you find integers x and y, where>> x^j = -y^j mod M>> where j is a positive odd integer, >> then it must be true that >> (x+y) = 0 mod f_1 or>> (x+y) = 0 mod f_2,>> if there doesn't exist integers c and d, such that M - c^j = d^j.>>But that also appears to be false. (As far as I can see neither 31*71>>nor 29*43 can be written as sum or difference of two cubes, the first>>can not be written as sum or difference of fifth powers and the second>>can not be written as sum or difference of seventh powers.)>>Where is your proof?> There may actually be an interesting theorem in here somewhere.> I found that, for j = 3, I could find lots of counterexamples> when f_1 and f_2 were primes congruent to 1 mod 6, but I couldn't> find any when either f_1 or f_2 or both were congruent to 5 mod 6.> Similarly, when j = 5, I could find counterexamples when > f_1 and f_2 were congruent to 1 mod 10, but not otherwise. Not> that I did really extensive searches. I don't see an immediate proof.I think that this follows from the fact that if j is relatively prime to p-1 then all numbers are jth residues mod p, so x^j == (-y)^j (mod p) => x == -y (mod p). I know that htis is handwaving, but I've a train to catch.> === Subject: Re: JSH: Basic result, test REVISION 2> Thanks to James Dolan for his examples withM=91, as apparently an> exception occurs with j=3, if there exists integers c and d, such that> M - c^3 = d^3.> You may also thank Nora Baron.> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2,> if there doesn't exist integers c and d, such that M - c^3 = d^3.> Let's see ... how about f_1 = 31, f_2 = 71, j = 5,> x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ?> Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ?> Or x = 91, y = 74 ? Or x = 95, y = 72 ?Looks like there's a size requirement, where in the lower rangesM^{1/j} needs to be greater than 5. That number may increase withincreasing j. === Subject: Re: JSH: Basic result, test REVISION 2>> Thanks to James Dolan for his examples withM=91, as apparently an>> exception occurs with j=3, if there exists integers c and d, such that>> M - c^3 = d^3.>> You may also thank Nora Baron.>> M = f_1 f_2>> if you find integers x and y, where>> x^j = -y^j mod M>> where j is a positive odd integer, >> then it must be true that >> (x+y) = 0 mod f_1 or>> (x+y) = 0 mod f_2,>> if there doesn't exist integers c and d, such that M - c^3 = d^3.>> Let's see ... how about f_1 = 31, f_2 = 71, j = 5,>> x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ?>> Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ?>> Or x = 91, y = 74 ? Or x = 95, y = 72 ?>Looks like there's a size requirement, where in the lower ranges>M^{1/j} needs to be greater than 5. That number may increase with>increasing j.Oh for heaven's sake will you ever shut up? This is mathematicswe're talking about. If you would say you think something mightbe true and ask whether people could settle the question thatwould be fine. But instead over and over you insist you canprove something, and every single time it turns out to be wrong.Hint: an actual _proof_ does not contain _guesses_.>************************ === Subject: Re: JSH: Basic result, test REVISION 2 > Thanks to James Dolan for his examples withM=91, as apparently an > exception occurs with j=3, if there exists integers c and d, such that > M - c^3 = d^3. > You may also thank Nora Baron. > M = f_1 f_2 > if you find integers x and y, where > x^j = -y^j mod M > where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or > (x+y) = 0 mod f_2, > if there doesn't exist integers c and d, such that M - c^3 = d^3. > Let's see ... how about f_1 = 31, f_2 = 71, j = 5, > x = 11, y = 9 ? Or x = 25, y = 17? Or x = 28, y = 65 ? > Or perhaps f_1 = 29, f_2 = 43, j = 7, x = 9, y = 7 ? > Or x = 91, y = 74 ? Or x = 95, y = 72 ? > Looks like there's a size requirement, where in the lower ranges > M^{1/j} needs to be greater than 5. That number may increase with > increasing j.Yeah, dream on. I thought you had a proof? If it really is a proof,and if there is a size requirement, you should carefully read theproof because that requirement ought to be present in your proof.-- === Subject: Re: JSH: Basic result, test REVISION 2... > M = f_1 f_2 > if you find integers x and y, where > x^j = -y^j mod M > where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or > (x+y) = 0 mod f_2, > if there doesn't exist integers c and d, such that M - c^j = d^j. > But that also appears to be false.A bit more thinking about it. Given x^j + y^j = 0 mod M with odd j.We can factorise (as Arturo also has done) as (x + y)(x^(j-1) - x^(j-2).y ... + y^(j-1)) = 0 mod M.For some reason James thinks that the second factor can not be divisibleby both f_1 and f_2. Nora's first example is revealing: f_1 = 31, f_2 = 71, j = 5, x = 11, y = 9.The factors are: x + y = 20and x^4 - x^3.y + x^2.y^2 - x.y^3 + y^4 = 11005.But 11005 = 31 * 71 * 5.So what is James' proof that the second factor can not be divided by bothf_1 and f_2? Note that there is no way to write 31*71 = 2201 as the sumor difference of n-th powers for whatever n > 2 you may like.(Of course, the reverse is true. Whenever there are c and d such thatM - c^j = d^j, c^j = -d^j mod M.)Conclusion: while James says he knows everything about algebraic integershe is not even able to get is stuff about plain integers right.-- === Subject: Re: JSH: Basic result, test REVISION 2> ...> M = f_1 f_2> if you find integers x and y, where> x^j = -y^j mod M> where j is a positive odd integer, > then it must be true that > (x+y) = 0 mod f_1 or> (x+y) = 0 mod f_2,> if there doesn't exist integers c and d, such that M - c^j = d^j.But that also appears to be false.> A bit more thinking about it. Given x^j + y^j = 0 mod M with odd j.> We can factorise (as Arturo also has done) as> (x + y)(x^(j-1) - x^(j-2).y ... + y^(j-1)) = 0 mod M.> For some reason James thinks that the second factor can not be divisible> by both f_1 and f_2. Nora's first example is revealing:> f_1 = 31, f_2 = 71, j = 5, x = 11, y = 9.> The factors are:> x + y = 20> and> x^4 - x^3.y + x^2.y^2 - x.y^3 + y^4 = 11005.> But> 11005 = 31 * 71 * 5.> So what is James' proof that the second factor can not be divided by both> f_1 and f_2? Note that there is no way to write 31*71 = 2201 as the sum> or difference of n-th powers for whatever n > 2 you may like.> (Of course, the reverse is true. Whenever there are c and d such that> M - c^j = d^j, c^j = -d^j mod M.)> Conclusion: while James says he knows everything about algebraic integers> he is not even able to get is stuff about plain integers right.Always tossing in the insults.It turns out that from what I've seen so far I *am* wrong with thisline of inquiry. That includes my last post where I considered a sizerequirement.But I want readers to consider the needling that usually goes alongwith posts by people like Dik Winter who have some weird agenda going.I think they're sadists or something.Oh well, I'm not terribly curious about such issues anymore. Thereare just all kinds of people out there, and it seems to me that Usenetis a natural place for certain types of people to take up residence. === Subject: Re: JSH: Basic result, test REVISION 2[snip]> It turns out that from what I've seen so far I *am* wrong with this> line of inquiry. That includes my last post where I considered a size> requirement.James, as has been no before, mathematics is simply too difficult for you to waste your time on. Youhave neither the skills for it nor the required ability to pay attention to detail. You seem to think thatsuccess in mathematics involves flashes of perceived brilliance which need to be shoved down the throatsof those less magnificent than yourself.Why not just accept your limitations and move on to something else better sui to you? Maybe a goodmatch would be standing outside McDonald's in a clown suit. Don't even bother with the suit, just beyourself.> --There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: Re: JSH: Basic result, test REVISION 2 > Thanks to James Dolan for his examples withM=91, as apparently an > exception occurs with j=3, if there exists integers c and d, such that > M - c^3 = d^3. > You may also thank Nora Baron.I was wondering a bit about whether Nora had pos it before James posthis. I found the following:James: time-stamp 16:51:21 (?) arrived here 02:00, localor 16:51:21 +0800. Anyhow, James pos this about 3 hours afterthe moment when Nora's refutation was available.he is in repeat mode of course.)-- === Subject: Re: JSH: Basic result, test REVISION 2> Wouldn't you know, I decided that it was wrong and saw what seems to> be an important condition I'd missed.Deja vu![snip]> I considered this to be a rather easy test, until I had problems> giving it!> What happend to YOUR proof????--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: Re: JSH: Basic result, test REVISION 2>I'd like to see someone prove that result, if you can.At least you changed the script enough so that it doesn't still say thatyou've proven it, and you'd like to see if anyone's as smart as you. === Subject: Factoring Conjecture: TripletsOne hallmark of new research is the ability to answer questions thatothers might not have even considered. I've been playing with aresult that I've mentioned before, but became curious enough to postit again, and see if I'm wrong in assuming that what's easy for me toprove, is not so easy for contemporary mathematicians usingcontemporary mathematics!I've pos that several times only to have to backtrack, and it maybe VERY annoying to some for me to say that, but the central pointthat new research brings up interesting new results remains.Hopefully I got it right finally.Following from my research on factoring polynomials intonon-polynomial factors I have a rather simple result that given primeintegef_1, f_2, and integer M, whereM = f_1 f_2if you find integers x, y and z, wherex^j = y^j = z^j mod Mwhere j is a positive integer greater than 1, then it must be true that x = y mod f_1 or x = z mod f_1, orx = y mod f_2 or x = z mod f_2.At this point I seriously doubt anyone can prove that conjecture withcontemporary mathematics, which is why I give it, as I think I canprove it.Then again, maybe not, but I'm not going to get emotional about it. === Subject: Re: Factoring Conjecture: Triplets> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, If it is easy for you to prove, your methods are sorelydeficient.> is not so easy for contemporary mathematicians using> contemporary mathematics! ... mainly because it's false ... > I've pos that several times only to have to backtrack, and it may> be VERY annoying to some for me to say that, but the central point> that new research brings up interesting new results remains.> Hopefully I got it right finally. Not yet.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it. You're partly right: no one can prove this conjecture withcontemporary mathematics.> Then again, maybe not, but I'm not going to get emotional about it. Why not? Most people would be very upset to find thatwhat they very confidently believed to be proofs were infact repealy wrong. They would start to wonder if theywere losing it, or if they don't really understand contemporarymathematics or what a proof actually is.> Again there is no shortage of examples: f_1 = 19, f_2 = 13, M = 247, j = 3: x = 89 y = 86 z = 72 x = 92 y = 81 z = 74 x = 99 y = 85 z = 63 x = 100 y = 92 z = 55 x = 106 y = 83 z = 58 x = 106 y = 96 z = 45 x = 109 y = 97 z = 41 etc., etc. (There are 259 examples like this with x, y & z <= 300.) Why you're doing here the last 2 days just looks like desperation. It's very easy to write a littleprogram to check on your conjectures for this kind of thing. Why not do that before you post? You just need the attention? === Subject: Re: Factoring Conjecture: TripletsDoes James have a job?> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> I've pos that several times only to have to backtrack, and it may> be VERY annoying to some for me to say that, but the central point> that new research brings up interesting new results remains.> Hopefully I got it right finally.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1,> then it must be true that> x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.> Then again, maybe not, but I'm not going to get emotional about it.> === Subject: Re: Factoring Conjecture: Triplets> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2Several people so far have pos counterexamples. Here's a whole class ofthem.Pick any two primes > 3 for these.> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1, Pick x=2, y=3, z=4, j=(f_1-1)(f_2-1).> then it must be true that > x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.x mod f_1 = 2, y mod f_1 = 3, z mod f_1 = 4x mod f_2 = 2, y mod f_3 = 3, z mod f_3 = 4-- === Subject: Re: Factoring Conjecture: Triplets> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> I've pos that several times only to have to backtrack, and it may> be VERY annoying to some for me to say that, but the central point> that new research brings up interesting new results remains.> Hopefully I got it right finally.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.> Then again, maybe not, but I'm not going to get emotional about it.> I thought about this:Let p be any prime, consider the arithmetic progressionn(p-1)+1now the initial term and the common difference are co-prime, so there areinfinitely many primes in this sequence aren't there? Dik, please can youput me right on that one if need be.Let one of these be q (q not equal to p).now, as p-1 divides q-1, by Fermat's little theorem for all non-zeroresidues mod p, a^(q-1) is congruent to 1 mod p and mod q. As p and q areco-prime, they are all congruent to 1 mod pq, right?thus there are p-1 distinct numbers mod p and q which have some power allcongruent to 1 mod pqp was arbitrary. so given any conjecture like James's latest one involvingan arbitrary number of elements x,y,z....w we can show there exists acounter example. Of course James might object because I can't produce analgorithm to give it to you explicitly.Yours,Matt Grime === Subject: Re: Factoring Conjecture: Triplets: Hopefully I got it right finally.: then it must be true that : At this point I seriously doubt anyone can prove that conjecture with: contemporary mathematics, which is why I give it, as I think I can: prove it.Dear JamesSince you are a moderately competent programmer, in the sense that yourcode runs, the proper approach to these sorts of conjectures, from anot-so-mathy point of view would be to write up a snippet of code whichtests lots of variables and see what comes up. Finding counterexamples to your *conjectures* takes seconds with a shortprogram.This way, you can keep your pseudoconjectures to yourself until you runinto something that actually *seems to hold* for a whole slew of cases,and *then* you can post it here and see what people think.Justin === Subject: Re: Factoring Conjecture: Triplets> : Hopefully I got it right finally.: then it must be true that : At this point I seriously doubt anyone can prove that conjecture with> : contemporary mathematics, which is why I give it, as I think I can> : prove it.> Dear James> Since you are a moderately competent programmer, in the sense that your> code runs, the proper approach to these sorts of conjectures, from a> not-so-mathy point of view would be to write up a snippet of code which> tests lots of variables and see what comes up. > Finding counterexamples to your *conjectures* takes seconds with a short> program.> This way, you can keep your pseudoconjectures to yourself until you run> into something that actually *seems to hold* for a whole slew of cases,> and *then* you can post it here and see what people think.> Justin(On the off chance the James reads this)You don't really need any code - thinking is just as quick for thesecases. Fermat's little theorem tells you thata^(p-1) is 1 mod p for p a prime, and all non-zero aall it takes is a couple of seconds to find two primes p and q where q-1divides p-1 and you can find q-1 distinct numbers (mod p and mod q) withtheir p-1th powers all congruent mod pq. === Subject: Re: Factoring Conjecture: Triplets> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> I've pos that several times only to have to backtrack, and it may> be VERY annoying to some for me to say that, but the central point> that new research brings up interesting new results remains.> Hopefully I got it right finally.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1,> then it must be true that> x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.> Then again, maybe not, but I'm not going to get emotional about it.> Good grief, James! How long are you going to persist in exhibiting yourincredible ignorance? You clearly haven't a clue about how mathematicsworks or what constitutes a proof. You are not only ineducable, butmilitantly so. (Damn the mathematicians! Full speed ahead!)Any sane person would have developed a complete mistrust in his ownflashes of insight or intuition, and learned how to construct a validproof. The time for posturing is over. Get a life.--There are two things you must never attempt to prove: the unprovable --and the obvious.----http://www.crbond.com === Subject: Re: Factoring Conjecture: Triplets>One hallmark of new research is the ability to answer questions that>others might not have even considered. I've been playing with a>result that I've mentioned before, but became curious enough to post>it again, and see if I'm wrong in assuming that what's easy for me to>prove, is not so easy for contemporary mathematicians using>contemporary mathematics!>I've pos that several times only to have to backtrack, and it may>be VERY annoying to some for me to say that, but the central point>that new research brings up interesting new results remains.There's nothing irritating about _that_ point - we all find theidea that it's somehow relevant to the things you post hilarious,a good time had by all. The irritating thing is the pompousbull about how you things that are easy for you are not soeasy for mathematicians. Given that you're wrong _so_often (including this time, it appears) you'd think you'dlearn to refrain from that sort of comment. Because you'dthink you'd eventually realize that it makes you sound likea total asshole.>Hopefully I got it right finally.Also a total idiot.>Following from my research on factoring polynomials into>non-polynomial factors I have a rather simple result that given prime>integef_1, f_2, and integer M, where>M = f_1 f_2>if you find integers x, y and z, where>x^j = y^j = z^j mod M>where j is a positive integer greater than 1, >then it must be true that >x = y mod f_1 or x = z mod f_1, or>x = y mod f_2 or x = z mod f_2.>At this point I seriously doubt anyone can prove that conjecture with>contemporary mathematics, which is why I give it, as I think I can>prove it.>Then again, maybe not, but I'm not going to get emotional about it.>************************ === Subject: Re: Factoring Conjecture: Triplets... > Hopefully I got it right finally.Why do you not first check whether what you have is right? It is soeasy to do. > Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integef_1, f_2, and integer M, where > M = f_1 f_2 > if you find integers x, y and z, where > x^j = y^j = z^j mod M > where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or > x = y mod f_2 or x = z mod f_2.f_1 = 7, f_2 = 13, M = 91, x, y, z any triple from{15, 18, 32, 44, 57, 58, 67} that does not contain one of the pairs(15, 57), (18, 32), (18, 44), (18, 57), (18, 67), (32, 58), (32, 67),(44, 57), (44, 58). So the triple (15, 18, 58) is one such triple. > At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it.Indeed... > Then again, maybe not, but I'm not going to get emotional about it.We'll see.-- === Subject: Re: Factoring Conjecture: Triplets> ...> Hopefully I got it right finally.> Why do you not first check whether what you have is right? It is so> easy to do.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, whereM = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1,> then it must be true that> x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> f_1 = 7, f_2 = 13, M = 91, x, y, z any triple from> {15, 18, 32, 44, 57, 58, 67} that does not contain one of the pairs> (15, 57), (18, 32), (18, 44), (18, 57), (18, 67), (32, 58), (32, 67),> (44, 57), (44, 58). So the triple (15, 18, 58) is one such triple.For, I suppose, j=3.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.> Indeed...> Then again, maybe not, but I'm not going to get emotional about it.> We'll see.Wow! Another blow to the very foundations of mathematics by James!He has discovered a theorem (it must be a theorem, otherwise he would nothave been able to construct a proof for it) to which you have, so it seems,been able to construct a counterexample!There must be something SERIOUSLY wrong with CORE MATHEMATICS!!-- Clive Toothhttp://www.clivetooth.dk === Subject: Re: Factoring Conjecture: Triplets> ...> Hopefully I got it right finally.Why do you not first check whether what you have is right? It is so> easy to do.Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, whereM = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1,> then it must be true that> x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.f_1 = 7, f_2 = 13, M = 91, x, y, z any triple from> {15, 18, 32, 44, 57, 58, 67} that does not contain one of the pairs> (15, 57), (18, 32), (18, 44), (18, 57), (18, 67), (32, 58), (32, 67),> (44, 57), (44, 58). So the triple (15, 18, 58) is one such triple.> For, I suppose, j=3.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.Indeed...Then again, maybe not, but I'm not going to get emotional about it.We'll see.> Wow! Another blow to the very foundations of mathematics by James!Nope. I'm just wrong. No big deal.> He has discovered a theorem (it must be a theorem, otherwise he would not> have been able to construct a proof for it) to which you have, so it seems,> been able to construct a counterexample!Irrational.> There must be something SERIOUSLY wrong with CORE MATHEMATICS!!You really need to get a life.I had an idea, tossed it out, and it didn't fly. No big deal.I wonder about some of you. === Subject: Re: Factoring Conjecture: Triplets> One hallmark of new research is the ability to answer questions that> others might not have even considered. I've been playing with a> result that I've mentioned before, but became curious enough to post> it again, and see if I'm wrong in assuming that what's easy for me to> prove, is not so easy for contemporary mathematicians using> contemporary mathematics!> I've pos that several times only to have to backtrack, and it may> be VERY annoying to some for me to say that, but the central point> that new research brings up interesting new results remains.> Hopefully I got it right finally.> Following from my research on factoring polynomials into> non-polynomial factors I have a rather simple result that given prime> integef_1, f_2, and integer M, where> M = f_1 f_2> if you find integers x, y and z, where> x^j = y^j = z^j mod M> where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or> x = y mod f_2 or x = z mod f_2.> At this point I seriously doubt anyone can prove that conjecture with> contemporary mathematics, which is why I give it, as I think I can> prove it.> Then again, maybe not, but I'm not going to get emotional about it.> how about x=1, y=2 z=4raise these to the 12th power and they are all congruent to 1 mod 7 andmod 13.