<s>[INST] How to prove that for all natural $n$, $133|11^{n+2}+12^{2n+1}$ 
How to prove that for every natural $n$, the number 
  $$11^{n+2} +12^{2n+1}$$
  is divisible by $133$?

I tried Induction method, so assuming that
$$
n=k \implies
A = (11^{k+2} +12^{2k+1} ) \pmod{133} \equiv 0
$$
Then trying to prove that
$$
n=k+1 \implies
B = (11^{k+3} +12^{2k+3} ) \pmod{133} \equiv 0
$$
For this I wanted to split $B$ into multiplication with at least one multiple of $A$.  But no luck.
Any Ideas?
 [/INST] $\displaystyle 11^{k + 3} + 12^{2k + 3} = 11 \times 11^{k + 2} +
\underbrace{\qquad 144\qquad}_{\displaystyle =\ 133 + 11} \times 12^{2k + 1} =
11\left(\color{red}{11^{k + 2} + 12^{2k + 1}}\right) + \color{red}{133} \times 12^{2k + 1}$
</s>