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?
 A: Notice $\ 133 =  11^2\!+\!11\!+1\ $ so we may apply
Lemma $ \bmod\ a^2\!+a+1\!:\,\ b := a^{n+2}\!+(a\!+\!1)^{2n+1}\!\equiv 0\,$ in any ring, $ $ for all $\,n\in\Bbb Z$
${\bf Proof}\,\ {\rm Note\ that}\ \  \color{#0a0}{a(a\!+\!1)\equiv -1}$ and $\,\color{#c00}{a^3\equiv 1}\ $ by $\,0\equiv (a\!-\!1)(a^2\!+a+1) = a^3\!-1,\,$ so
$\ a^{2n+1}b\, =\, a^{3n+3}\! + (\color{#0c0}{a(a\!+\!1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0\ $ so $\ b\equiv 0,\,$ by $\,a$ $\rm\color{#c00}{unit}$ so cancellable.

Or use $\ \color{#90f}{a\!+\!1}\equiv -a^2\,\Rightarrow\, (\color{#90f}{a\!+\!1})^{2n+1}\equiv -a^{4n+2}\equiv -(\color{#c00}{a^3})^n a^{n+2}\equiv -a^{n+2}$
Remark $ $ Both are essentially special cases of the method used here, which is a special case of the method of simpler multiples.
A: See that $12^2 = 144 \equiv 11 \pmod{133}.$
So we have 
$$11^{n+2} + 12^{2n+1} \equiv 121(11)^n +12(144)^n $$
$$\equiv 121(11)^n  +12(11)^n \equiv 133(11)^n \equiv 0 \pmod{133}.$$
A: 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
$$11^{n+1} + 12^{2n+1}  \equiv 19\cdot11^n\equiv 0 \bmod 19$$
