mm-169
Given integer n>=1 let a(n) = number of non-congruent solutions
ofx^3 + y^3 == 1 mod n .Is there a simple formula for a(n)?
Given integer n>=1 let a(n) = number of non-congruent
solutions of> x^3 + y^3 == 1 mod n .> Is there a simple
formula for a(n) ?Theorem 2 of Section 3 of Chapter 8 of
Ireland and Rosen, A Classical Introduction to Modern Number
Theory: Suppose p is a prime, congruent to 1 mod 3. Then there
are integers A and B such that 4p = A^2 + 27 B^2. If we require
A congruent 1 mod 3, A is uniquely determined, and the number
Daniel Grayson, dan@math.uiuc.edu, moderator for
sci.math.research
=>Given integer n>=1 let a(n) = number of
non-congruent solutions of>x^3 + y^3 == 1 mod n .>Is there a
simple formula for a(n) ?This is sequence A087412 in the
On-line Encyclopedia of Integer
Sequences<http://www.research.att.com/~njas/sequences/
Seis.html>. No formulafor it is given there. Note however that
if phi(n) is not divisible by 3, the map x -> x^3 is 1-1 and
onto on Z_n, so in that case a(n) = n. Also, a(n) is a
multiplicative function, i.e. a(mn) = a(m) a(n) if m and n
arerelatively prime.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
Grayson, dan@math.uiuc.edu, moderator for
sci.math.research
=>Given integer n>=1 let a(n) = number
of non-congruent solutions of>x^3 + y^3 == 1 mod n .>Is
there a simple formula for a(n) ?> This is sequence A087412 in
the On-line Encyclopedia of Integer Sequences>
<http://www.research.att.com/~njas/sequences/Seis.html>. No
formula> for it is given there. Note however that if phi(n) is
not divisible by 3, > the map x -> x^3 is 1-1 and onto on Z_n,
so in that case a(n) = n. Oops: of course it's not true that 
x
-> x^3 is 1-1 and onto onZ_n. It's 1-1 and onto on the group
G(n) of members of Z_n relatively prime to n. Nevertheless, I
believe it is still true that a(n) = n if phi(n) isnot
divisible by 3. Moreover (from numerical evidence) it looks
likea(p^k) = p^(k-1) a(p) for any prime p <> 3 and positive
integer k, while a(3^k) = 3^(k-2) a(9) for k >= 2.Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
=Your conjectural formula
a(p^k) = p^(k-1) a(p) for any prime p <> 3 and positive
integer kis correct. This follows from Hensel's lemma, since
the equationx^3+y^3=1 is non-singular modulo p, except for
p=3. Your formula whenp=3 is also 'ne; Hensel's 
lemma says
that even if the equation issingular, you simply need to start
with some higher power congruence,depending on how bad the
singularity is.In general, if a(f,n) denotes the number of
solutions to f(x,y)=0,where f(x,y) is a polynomial with
integer coef'cients, then a(f,p^k) = p^{k-1}a(f,p) provided
the curve is nonsingular, or what comes to the same
thing,provided f, df/dx, and df/dy do not have a simultaneous
root in thealgebraic closure of Z/pZ. In the present instance,
f = x^3+y^3-1, df/dx = 3x^2, df/dy = 3y^2,so it's clear 
there
is no simultaneous solution unless p = 3.Suppose now that
phi(n) is not divisible by 3. Since both phi(n) anda(n) are
multiplicative (Chinese remainder theorem), it suf'ces 
toprove
that a(n) = n when n is a power of a prime, say n =
p^k.Further, from the formula above, unless p = 3, it's 
enough
to check itfor n = p. But phi(p) = p-1, so it suf'ces to 
check
it when p-1 isnot divisible by 3. In other words, it needs to
be checked when p = 2(mod 3), and you already noted that it is
true in that case sinceevery element of Z/pZ is a cube.That
leaves the case p=3 to check. But if n = 3^k with k > 1,
thenphi(n) is divisible by 3. So we're left to check that 
a(3)
= 3, andthe solutions mod 3 are (1,0), (0,1), (2,2). That
completes the proofthat a(n) = n when phi(n) != 0 (mod
3).Finally, to get back to the question of the value of a(p),
the hardcase is p = 1 (mod 3). In that case, there are
formulas for a(p) usingcubic residue symbols (see, e.g.,
Ireland-Rosen's book on numbertheory). More importantly, if
you build all of the a(p) values into anL-series, it turns out
that the L-series factors as a product of twoHecke L-series.
This is a particular case of the theory of
complexmultiplication of elliptic curves.JHS>Given integer
n>=1 let a(n) = number of non-congruent solutions of>x^3 +
y^3 == 1 mod n .>Is there a simple formula for a(n) ? This is sequence A087412 in the On-line Encyclopedia of
Integer Sequences>
<http://www.research.att.com/~njas/sequences/Seis.html>. No
formula> for it is given there. Note however that if phi(n)
is not divisible by 3, > the map x -> x^3 is 1-1 and onto
on Z_n, so in that case a(n) = n. > Oops: of course it's not
true that x -> x^3 is 1-1 and onto on> Z_n. It's 1-1 and 
onto
on the group G(n) of members of Z_n relatively prime > to n.
Nevertheless, I believe it is still true that a(n) = n if
phi(n) is> not divisible by 3. Moreover (from numerical
evidence) it looks like> a(p^k) = p^(k-1) a(p) for any prime p
<> 3 and positive integer k, while > a(3^k) = 3^(k-2) a(9) for
k >= 2.> Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel > University of
moderator for sci.math.research
=[A complimentary Cc of
this posting was sent toLaurent Demanet > Consider the
eigenvalue problem A R = R Lambda, where A is a n-by-n >
matrix of operators acting on functions.Forget about
functions; simplify the stuff, and assume that entries inA are
m x m matrices.> We seek eigenvectors R_i and eigenvalues
lambda_i that are> themselves operators.Not clear what you
mean. Should R be an n-vector with entries beingoperators?
Then, essentially, you want to solve A R = R Lambdawith A
being a given nm x nm matrix, R being an unknown nm x m
matrix,and lambda being an unknown m x m matrix? Then the
problem reduces to'nding m-dimensional invariant subspaces of
A - which is easy if youknow diagonalization (or the Jordan
form) of A.[The answer is essentially the same in
in'nite-dimensional case - youshould look for closed 
invariant
subspaces of in'nite dimension andcodimension.]I know some
other interpretations of your question - and they
havedifferent answers...Hope this helps,Ilya
Not clear
what you mean. Should R be an n-vector with entries being>
operators? Then, essentially, you want to solve> A R = R
Lambda> with A being a given nm x nm matrix, R being an
unknown nm x m matrix,> and lambda being an unknown m x m
matrix? Then the problem reduces to> 'nding m-dimensional
invariant subspaces of A - which is easy if you> know
diagonalization (or the Jordan form) of A.> [The answer is
essentially the same in in'nite-dimensional case - you> 
should
look for closed invariant subspaces of in'nite dimension and>
codimension.]the full diagonalization is a particular instance
of an incomplete one. There are plenty of ways to identify any
number of in'nite-dimensional subspaces from the spectral
representation of A. My question was poorly phrased. What I
had in mind were examples like the following one :(A f)(x) =
(0, d/dx (c(x)*f(x)) ; - c(x)*d/dx (f(x)) , 0),which can be
diagonalized as A R_pm = R_pm Lambda_pm
withLambda_pm^2 f = -c d^2/dx^2 (c*f),take the 2 square
roots (say |log c| is bounded and smooth), andR_pm f = (
d/dx (c*f) , Lambda_pm f ).I was wondering if this
particular construction was part of a broader calculus. I
think it's called a one-way wave'eld 
decomposition.> I know
some other interpretations of your question - and they have>
for sci.math.research
=The following paper has been
published:Algebraic and Geometric
TopologyURL:http://www.maths.warwick.ac.uk/agt/AGTVol3/agt-3-
38.abs.htmlTitle:Addendum to Coarse homology
theoriesAuthor(s):Paul D. MitchenerAbstract:theories
Numbers. Primary: 55N35, 55N40Secondary: 19K56,
46L85Keywords:Coarse geometry, exotic homology, coarse
Baum-Connes conjecture, Novikov conjectureReceived: 12
September 2002Author(s) address(es):Institut fuer Mathematik,
Universitaet GoettingenD-37083 Goettingen, GermanyURL:
dan@math.uiuc.edu, moderator for sci.math.research
=>Given
is a cube in R^K> Q = { x is element of R^K | 0 <= x <= 1
}>and a linear map> A : R^K -> R^2.>The image of the cube A(Q)
is a convex polyhedron which can be described by>a system of
linear inequalities> A(Q) = { y is element of R^2 | B*y <= C
}>with a (m,2)-matrix B and a (m,1)-vector C.>The problem is
to 'nd (m,B,C) as a function of the linear map A.Let A_n be
the image of {y in Q: y_j = 0 for j > n}. So each A_n is a
convex polygon, A_2 being a (possibly degenerate)
parallelogram,and A_{n+1} is the convex hull of A_n and its
translate by A(e_{n+1})where e_n is the vector in R^K with 1
in position n and 0 elsewhere.Going from A_n to A_{n+1} is
easy: for each segment s of the boundary of A_n where the dot
product of A(e_{n+1}) and the outward-pointing normal to s is
negative, you take s; for each segment s where the dot product
is positive you take s + A(e_{n+1});and for the two points p
where the dot product changes sign youtake segments p to p +
A(e_{n+1}).(Modify this slightly to take care of the case
where a dot product is 0)Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
=Informal Inductive reasoning
seems to imply the image in R^2 has 2kcorners (or less).1) Is
this true?2) Is there a similar formula for R^3?3) For what
more general class of 'gures than a k-cube does 
thishold?FIRST
name on the list (No.1) along with a note in the NOTE