Prove that if $7$ divides $6^n + 1$ then $n$ is odd Prove that if $7$ divides $6^n + 1$ then $n$ is odd
Attempt:
We'll prove the contrapositive: $n$  is not odd if $7$ does not divide $6^n + 1$ 

If $x+1$ divides $x^n+1$ then $-1$ is a root of $x^n+1$. Therefore $(-1)^n+1=0$ which is only possible if $n$ is odd. Now put $x=6$.
