From: Fred W. Helenius <fredh@ix.netcom.com>
Subject: Re: Fermat-like equation in GL(k,Q)
Date: Sat, 15 Jan 2000 06:25:05 -0500
Newsgroups: sci.math
Summary: [missing]

David Bernier <bernier@my-deja.com> wrote:

>In article <85oap6$m1e$1@nnrp1.deja.com>,
>  David Bernier <bernier@my-deja.com> wrote:
>> FLT can be reformulated as:
>>``If A,B and C are in GL(1,Z) and n=>3, then A^n+B^n+C^n is not 0." (*)

>I think the amended statement obtained from the (*) above by
>replacing 'Z' by 'Q' is right since the equation is homogeneous
>of degree n.

OK.  Now I'm with you.

>  Below I reformulate my general question:

> The general question is obtained by letting k be arbitrary, i.e.

> ``For a given k=>2, for which values of n=>2 is it the case that
>   the matrix equation A^n +B^n +C^n = 0 has no solutions where
>   A, B and C are all in GL(k,Q)?"

A 2x2 example (which can, of course, be extended to kxk by putting
ones on the diagonals) serves to show that there are solutions for
all odd n:

Let A, B, C be

[0  1]  [-1 1]  [ 1  0]
[-1 0], [ 0 1], [-1 -1];

then A^3 + B^3 + C^3 = (-A)^5 + B^5 + C^5 = 0, and similar equations
hold for higher odd powers since A^4 = B^4 = C^4 = I.


Similarly, let A, B, C be

[ 0  1]  [-1 -1]  [1  0]
[-1 -1], [ 1  0], [0  1];

then A^n + B^n + C^n = 0 whenever n is not a multiple of 3
(A, B, C are cube roots of I).


The remaining values of n, the multiples of 6, seem to be harder.

--
Fred W. Helenius	<fredh@ix.netcom.com>
==============================================================================

From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: Fermat-like equation in GL(k,Z)
Date: 17 Jan 2000 05:04:07 GMT
Newsgroups: sci.math

In _13 Lectures on Fermat's Last Theorem_ chapter XIII section 9
Ribenboim mentions results about solutions of A^n+B^n=C^n in
square matricies A, B, C. One partial result restricting the solutions
is:

   Theorem (Barnett and Weitkamp). Let n>2, n<>4. If Fermat's theorem
   holds for the exponent n (when n is odd) or for the exponent n/2
   (when n is even) if A, B are nonsingular 2x2 matricies over Q,
   which are not scalar matricies but such that A^n, B^n are
   scalar matricies, then there exists no 2x2 nonsingular matrix C,
   with entries in Q, such that A^n+B^n=C^n.

They also have counterexamples with nonsingular matricies:

For n=1(mod 4),

A = (-1 -1)     B = ( 1  2)   C = (-2 -3)
    ( 0  1)         (-1 -1)       ( 1  2)

For n=3(mod 4),

A = (-3 -5)     B = ( 1  2)   C = (-2 -3)
    ( 2  3)         (-1 -1)       ( 1  2)

These matricies are fourth roots of the identity matrix.

Keith Ramsay