[levn11]

https://www.techrxiv.org/users/717330/articles/702287-on-fer...

[n4r9]

On p.4 you argue that for integers a, b, c and n:

  (a + b − c)^n = (c − a)(c − b)g_1(n)

  => a + b − c = [(c − a)(c − b)g_1(n)]^(1/n)

  => g_1(n) | a + b - c

This doesn't follow as it stands. For example, if a=b=3 and c=n=2, then g_1(n)=16 whereas a + b - c = 4.

[levn11]

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.

[n4r9]

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.

[levn11]

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.

[n4r9]

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.

[levn11]

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.

[levn11]

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.