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Comment by levn11 | Hacker News Reader

This is only true for a,b,c that satisfies a^2+b^2=c^2.

So a=b=3 and c=n=2 is not part of the solution set.

But you're not using a^n+b^n=c^n in your argument at the bottom of p.4. You just say "therefore g_1(n) divides a+b-c". My example shows that yhe implication doesn't follow in general. And it's not clear why it follows specifically if a,b,c are a solution to FLT.

i believe you skipped pages 1-3. g_1(2)=2 for all a,b,c with n=2. g(n) carries with it the assumption of a^n+b^n=c^n as i showed in pages 1-3.

Pages 1-3 simply show that (c-a) and (b-a) divide (a+b-c)^n (for even n), assuming they are a solution to FLT.

You then define g to be (a+b-c)^n/(c-a)(b-a), an integer.

I follow you this far. I do not see why g divides a+b-c, and I don't think the argument on p.4 proves it.

I won't reply further to this question about g, i do think i've been clear. and at this point you can be on your merry way still thinking it's wrong. but you simply misunderstood.

I've read your paper and followed the arguments and this is where I believe it falls down. Either that or it needs much better explanation.

it's a proof by contradiction. g would divide a+b-c IF a+b-c are integers.

for n=2, g(2)=(c-a)(c-b)g_1(2) and g_1(2)=2.

So only when n=2 is it true that g divides a+b-c.

Otherwise we get a contradiction that it divides. since then, g_1(n) for n>2 is not a factor of a+b-c, we can safely assume at least one of them was not an integer.

I honestly don't follow your last sentence. Why does g not being a factor of a+b-c mean they're not integers?

it follows specifically from the form on pages 1-3. i would recommend reading it with fresh eyes after a good night's rest.

It doesn't follow from anything on p1-3. Certainly not directly. If you were being genuine about this I think you would appreciate an opportunity to improve the proof rather than resort to insults!

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