[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.

[n4r9]

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

[levn11]

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

[n4r9]

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!