I was asked to prove : Let k be a positive integer. Prove that gcd(k, 2^n + 3^n + 6^n - 11) = 1 for every integer n >= 2 if and only if k = 1.
I tried to use FlT, 2^p congruent to 2, 3^p congruent to 3, and 6^p congruent to 6, then 2^n + 3^n + 6^n - 11 congruent to 2 + 3 +6 -11 = 0 so a prime divides the 2^n + 3^n + 6^n - 11, as we have gcd(k, 2^n + 3^n + 6^n - 11) = 1, so k must be 1.
Then I found there's a mistake there, as 6 is not prime, we can not say 6^p congruent to 6 (mod 6).
But seems my direction is still correct. Any idea to improve it? Thanks a lot!
Suppose $p|k$, $p$ prime. Set $n=p$. Then by LFT we have $2^n+3^n+6^n-11\equiv 2+3+6-11\mod p$, which is $0$. Thus $p$ also divides this expression. So if for any $n$ we have $\gcd(\ldots)=1$ then $k$ must not be divided by any prime $p$, hence $k=1$. The other direction is trivial.
LFT works for any integer number, it says: If $a\in\mathbb Z$ and $p$ prime then $a^p\equiv a\mod p$. So it only requires $p$ to be prime, $a$ can be any number.