Re: Even Periods in Fermat's Last Theorem and Modular Arithmetic (FLTMA)


quasi wrote:
> On 31 Jul 2005 18:41:22 -0700, DGoncz@xxxxxxx wrote:
>
> >I wrote:
> >0<n<a<b<c<(a+b)
> >
> >I meant:
> >0<n,a,b,c;a<b<c<(a+b)
> >
> >Doug
>
> Try testing one aspect of your idea for n=3.
>
> See if you can find integers a,b,c satisfying your inequalities and
> also the 3 congruences:
>
> a^3+b^3=0 mod c
> c^3-a^3=0 mod b
> c^3-b^3=0 mod a

Oh, I have tried:
for c = 5 to limit
for b = 4 to c-1
for a = 3 to b-1
if (a<b<c<(a+b)) etc

to limits like 13 or 101 (hypotenuses of Pythagorean triangles)
and have noted that with the inequality, the exponent is always even,
while without the inequality, the exponent can be odd.

>
> If you find them, wouldn't that put you back to square 1?

Yes, it would, but it doesn't. That is, I haven't found one.

I could increase the limit and only calculate the congruences to n = 3.
That might be fun.

> By the way, did you ever use the inequalities a<b<c<a+b? If it plays
> no role in your proof, why even state it?

See the above.

> Finally, if you find solutions to the above congruences, check if in
> all cases. s.n=0 mod abc. If even one case fails, then you can't
> assert s.n = 0 mod abc.

I've extended that notion in a post farther down this thread. It was
wrong as first stated.

>
> The case n=3 is a good test case for proposed elementary attacks on
> FLT. If it doesn't work for n=3 then the attack is flawed, and if it
> does work, then that's already a worthwhile discovery since, as far as
> I know, there are no known elementary proofs of FLT for n=3 (all use
> algebraic number theory).
>
> quasi

I guess so. It's all about n being even and prime. I have not stated
the arguments for n prime. I read about it in Fermat's Last Theorem for
Amateurs.

Thanks so much for participating.

Doug

.

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