Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$ Prove $133\mid \left(11^{n+2}+12^{2n+1}\right)$, where $n$ is a non-negative integer.
So, I went about proving this using Fermat's theorem. 
So I had $11^{n}\cdot 11^2+(12^2)^n= 0\ (\mod 133)$
then $11^n\cdot 11^2+1728^n=0\ (\mod133)$ 
and finally $1^n+132^n=0\ (\mod 133)$. 
Then I said $133^n=0\ (\mod133)$ because $133$ raised to any power will have a remainder of $0$. Did I apply Fermat's theorem correctly?

since $133=7*19$, first show that $$7\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$
then $$19\, |\, \left(11^{n+2} + 12^{2n+1}\right)$$
Since $gcd(7,19)=1$, the result follows.
$$ 11^{n+2}\equiv 11^{n}\cdot 11^2\equiv 2\cdot11^{n}\bmod 7$$
$$12^{2n+1}\equiv 5\cdot 144^{n}\equiv 5\cdot 11^{n}\bmod 7$$
So $$11^{n+1} + 12^{2n+1}  \equiv 7\cdot11^n\equiv 0 \bmod 7$$
$$..........................................................................$$
$$11^{n+2}\equiv 121\cdot 11^n\equiv 7\cdot11^{n}\bmod 19$$
$$12^{2n+1}\equiv 12\cdot 144^{n}\equiv 12\cdot 11^{n}\bmod 19$$
So
$$11^{n+1} + 12^{2n+1}  \equiv 19\cdot11^n\equiv 0 \bmod 19$$
