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