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?

Given that $11^{k+2}+12^{2k+1}=133\lambda$, you can infer that $11^{k+2}=133\lambda-12^{2k+1}$.  
Thus you have that $$11^{k+3}+12^{2k+3}= 11^{k+2}\cdot 11 +144 \cdot 12^{2k+1}=11\cdot 133\lambda -11\cdot 12^{2k+1} +144 \cdot 12^{2k+1} =11\cdot 133\lambda +12^{2k+1} \cdot (144-11)=11\cdot 133\lambda +133\cdot 12^{k+1} $$
 which is divisible by 133
