Re: Computability

On Thu, 10 Jan 2008 06:15:55 -0500, quasi <quasi@xxxxxxxx> wrote:

On Wed, 9 Jan 2008 09:24:00 -0800 (PST), The Dougster 22044
<DGoncz@xxxxxxxxxxxx> wrote:

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

It looks like the above line is not used again.

for x,y,z in Z and p in P (the Primes), x^p + y^p = z^p

So the above line is your true starting point.

You forgot to mention the standard restriction

x*y*z is nonzero

Try to be more careful -- if there are restrictions, declare them.

The word "and" in the line below is not correct.

Presumably, you meant "then".

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,

I broke the above into 4 separate lines to improve readability.

But already there are some unjustified conclusions.

To get from

x^p + y^p = 0 mod z

to

(x/y)^p = -1 mod z

you need to have (y,z) = 1.

Did you mention that restriction anywhere? No.

Ok, so let's add that.

We can assume, without loss of generality, that x,y,z are pairwise
coprime.

Now your conclusions above are ok.

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"),

In the above 2 lines you are defining the notation o(a,m).

Fine.

o(x/y, z) = o(y/x, z) = 2p and likewise

The above line above needs more work.

How do you know that o(x/y, z) is not equal to 2?

I can see how to prove it, but it requires some discussion.

o(z/x, y) = o(x/z, y) = p and
o(z/y, x) = o(y/z, x) = p.

The 2 lines above are ok.

Also, x<y<z<(x+y) by a theorem of Mahanobolis.

Firstly, you never declared x,y,z are _positive_ integers. All you
said was that x,y,z are in Z. There's nothing wrong with assuming
x,y,z are all positive integers -- my complaint is that you never
_mentioned_ it.

Same thing with regard to the inequality x < y. You can't _prove_
that. However by symmetry, you can always _assume_ x <= y.

Given that x,y,z are now assumed as positive integers,

proving y < z is trivial,

and proving x not equal to y is easy.

It's also easy to prove z < x + y, hence you get

0 < x < y < z < x + y,

as you had claimed.

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)

Fine, provided you have shown o(x/y, z) is not equal to 2, but that
_can_ be shown, so your conclusion is ok.

and existing proofs of FLT show us p = 2 for all solutions to FLT,
but what of the converse?

The converse to what statement?

Are there any candidate triples with p>2?

Apparently, the above is what you mean.

I say there are none.

Interesting conjecture.

I find in tests to z = 31, 63, 127, or 256 that there are none there.

Ok.

Chip Eastham has suggested a test to z = 65535

A brute force search might not be able to reach that high in any
reasonable amount of time.

would be suggestive

Suggestive of the truth of the conjecture? Maybe.

and might inspire a proof.

That seems unlikely, unless by "inspire" you mean that if the search
comes up empty, that provides some confidence that it's worth trying
to _find_ 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.

That's a cool conjecture.

(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.

You did pretty well -- much better than last time.

The flaws of your exposition (as noted in my nitpicks) are minor, and
easily fixed.

But flaws aside, the essence of your claim is now very clear.

Thanks for your efforts to improve the explanation.

But your search program needs some work.

You claimed to have checked it up to z = 256, but apparently you
missed the following counterexample:

(x,y,z) = (74,129,143) which has signature (6,3,3).

quasi
.

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