# A Proof of Fermat's Conjecture

#### skipjack

Forum Staff
How many previous attempted proofs have you devised? How many will you need to make before giving up?

#### SDK

How many previous attempted proofs have you devised? How many will you need to make before giving up?
The worst part is to think of all the ACTUAL mathematics these people could be doing with the same time and effort they expend on trying to prove FLT with elementary algebraic manipulations, disprove Cantor, or prove that $.999... \neq 1$, etc etc. Such a waste.

#### bruno59

Unlike skipjack and SDK, I will not judge before seeing.
So I saw and I'm interested. Because your work, my dear Reyenshun, reminds me of mine.
For a while, I've been up in the case p|abc
I used your x,y,z too, which I called remainders and I also proved your x+y-z=0 (1),
not by some "foggy" way, but by showing that
p$$\displaystyle ^2$$|x$$\displaystyle ^{p-1}$$-y$$\displaystyle ^{p-1}$$
Eventually, I ended in
x$$\displaystyle ^p$$+y$$\displaystyle ^p$$-z$$\displaystyle ^p$$, which contradicts (1) of course.
But I wouldn't post anything before addressing case p|abc too. Now I see I might change my mind.
As for your attempt, I've spotted at least one flaw, fatal or not I don't know.
I'll have a second glance when I'll be able.

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#### SDK

Unlike skipjack and SDK, I will not judge before seeing.
So I saw and I'm interested. Because your work, my dear Reyenshun, reminds me of mine.
There is a simple theorem which proves that algebraic manipulation is incapable of producing a proof of FLT. Not "we haven't found one yet" or "maybe if I am smarter than anyone else who tried...", but rather, it's a THEOREM that it can't be done.

This is because any supposed proof based on algebraic manipulations (multiplication, addition, taking powers, inverses) would also hold equally well over the p-adics for p prime. But FLT has solutions over the p-adics for EVERY prime.

Thus, there is no reason to go diving into his supposed proof to find the mistake. That doesn't mean I'm not confident it's there.

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• 1 person

#### retenshun

three or four, perhaps.

#### retenshun

Unlike skipjack and SDK, I will not judge before seeing.
So I saw and I'm interested. Because your work, my dear Reyenshun, reminds me of mine.
For a while, I've been up in the case p|abc
I used your x,y,z too, which I called remainders and I also proved your x+y-z=0 (1),
not by some "foggy" way, but by showing that
p$$\displaystyle ^2$$|x$$\displaystyle ^{p-1}$$-y$$\displaystyle ^{p-1}$$
Eventually, I ended in
x$$\displaystyle ^p$$+y$$\displaystyle ^p$$-z$$\displaystyle ^p$$, which contradicts (1) of course.
But I wouldn't post anything before addressing case p|abc too. Now I see I might change my mind.
As for your attempt, I've spotted at least one flaw, fatal or not I don't know.
I'll have a second glance when I'll be able.
How does x^p + y^p - z^p contradict x + y - z = 0?

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#### bruno59

How does x^p + y^p - z^p contradict x + y - z = 0?
Looks like I can't write down a sentence without a typo
I meant that $$\displaystyle x^p+y^p-z^p=0$$,which I ended to, contradicts x+y-z=0

Ignoring for now that there is no explicit proof of x+y+z=0, can you please enlighten me about how (case $$\displaystyle p\nmid abc$$):
$$\displaystyle (x^p+y^p)/(x+y)$$=$$\displaystyle \prod_{r=1}^u(pQ_r+1)^{p-1}$$