Re: Computability

On Jan 9, 6:23 am, quasi <qu...@xxxxxxxx> wrote:
...
A few simple numerical examples might have made up for the coded
wording. Your original post suggests that you have such examples. Why
did you not bother to share them? Remember -- one writer, multiple
readers. A little extra effort on your part (one person) benefits all
the readers (many people). Empathize with the many, and be generous
with your explanatory efforts (within reason).

Ok, that's my critique.

Perhaps it will help you improve your writing style.

OK.

If
for a,b,c, and n in Z, a^n + b^n = c^n then
for x,y,z in Z and p in P (the Primes), x^p + y^p = z^p and
x^p + y^p = 0 mod z, x^p / y^p = -1 mod z, (x/y)^p = -1 mod z, and (y/
x)^p = -1 mod z, where
/y or /x is the multiplicative inverse of y or of x, so
with o(a,m) = the order of a in m with (a,m), the gcd of a and m = 1
(a and m are "coprime"),
o(x/y, z) = o(y/x, z) = 2p and likewise
o(z/x, y) = o(x/z, y) = p and
o(z/y, x) = o(y/z, x) = p.
Also, x<y<z<(x+y) by a theorem of Mahanobolis.

The "signature" of a "candidate triple" x,y,z, with is then:

( o(x/y, z), o(z/x, y), o(z/y, x) ) = ( 2p, p, p) and
existing proofs of FLT show us p = 2 for all solutions to FLT, but
what of the converse?
Are there any candidate triples with p>2?

I say there are none.

I find in tests to z = 31, 63, 127, or 256 that there are none there.
Chip Eastham has suggested a test to z = 65535 would be suggestive and
might inspire a proof.

I note the signature of
x y z is:
3 4 5 4 2 2
5 12 13 4 2 2
7 24 25 4 2 2
13 15 16 4 4 4 with x < y < z < x+y and (x,y)=(y,z)=(z,x)=1
19 45 46 6 6 6 ditto

So, restating as precisely as I am able today,

If x < y < z < x+y and x,y,and z are pairwise coprime, and
o(x/y, z), o(z/x, y), o(z/y, x) = 2p, p, p (prime p)
then p = 2.

(I'd previously in this thread forgotten about that -1 and that 2p.
Sorry. I have it right on my resume.)

Now, knowing FLT true does NOT prove this proposition. Knowing this
proposition true does, however, prove FLT by a new route, one not
involving modular curves, as it would rule out all of what I call
"candidate triples" (x,y,z) for which one might check to see if x^p +
y^p = z^p in fact.

And that's about the best I can do today. My cat is in my lap, and
needs a pet, so I'll check the thread tomorrow or maybe at best later
today.

Thanks all.

Doug
.

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