Re: FLTMA: Verification of "large zero"

DGoncz_at_aol.com
Date: 03/18/05 

Date: 18 Mar 2005 05:24:39 -0800

richard miller wrote:
> <DGoncz@aol.com> wrote in message
> news:1110923062.398760.6430@o13g2000cwo.googlegroups.com...
> > Can any reader here verify that
> >
> > for n=546, a=49, b=51, and c=53,
> >
> > (a^n + b^n) mod c =
> > (c^n - a^n) mod b =
> > (c^n - b^n) mod a = 0?
>
> So, with all the good replies in the bag (esp. the terse, but neat
reply
> from Hanford Carr) one might well wonder where Doug Goncz dreamt up
such a
> ludicrous exponent).
>
> The clue lies in factoring 546 = 2 * 3 * 7 * 13
>
> Now there's a thing!
>
> My money is on Doug studying cubic exponent residues, i.e. x^3 mod A,
where
> A is 49, 51 or 53.

No money. I wasn't looking at cubic exponent residues at all.

I have been looking at the series

S.n=
(a^n + b^n) mod c +
(c^n - a^n) mod b +
(c^n - b^n) mod a

and wondering if its lattice representation tests out differently for
those series having a zero.

I have been looking at George Marsaglia's classic paper on linear
congruential sequences, and have programmed a few of the definitions
and functions in Mathcad.

In this case S(49,51,53) has a zero S.546.

This is related to Fermat's Last Theorem, and FLTMA stands for Fermat's
Last Theorem and Modular Arithmetic, a flag or tag I propsed a while
ago here.

I admit

a^n == -b^n mod c or mod c^n

is kind of intruiging.

The most important thing, though is that 546 is even. If it were
possible to prove that S.n = 0 ==> n == 0 mod 2, then that would open
up an exploration into FLT.

The additional conditions are only 0 < a < b < c < (a+b).

Can anyone here verify as expected that

49^546 + 51^546 - 53^546? != 0?

Yours,

Doug Goncz
Replikon Research
Seven Corners, VA 22044-0394

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