-6 karma · joined November 4, 2023
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.
So a=b=3 and c=n=2 is not part of the solution set.
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If you would like to check out an elementary proof for Fermat's Last Theorem (or at least an attempt)