$$133|11^{n+1} + 12^{2n-1}$$
I proof by Induction
Basic step
$$
\begin{align}
P(1); 133|(121 + 12)
\end{align}
$$
Inductive step
$$ \begin{align} P(k);&133|(11^{k+1}+12^{2k-1})\\ \\ P(k+1); & 11^{k+2} +12^{2k+1}\\ &(12-1)11^{k+2} + 12 ^{2k+1}\\ &12\cdot 11^{k+1} + 12 ^{2k + 1}\\ &12(11^{k+1}+12^{2k}) -11^{k+1}\\ &12(11^{k+1}+12^{2k-1}+11\cdot12^{2k-1})-11^{k+1}\\ &12(11^{k+1}+12^{2k-1}) + 11(12^{2k}-11^k) \end{align} $$ I can't solve more, please help!
It's wrong! Try $n=1$, for which we need $1849$ is divisible by $133$.
If you mean to prove that $$11^{n+2}+12^{2n+1}$$ is divisible by $133$ then we have: $$11^{n+2}+12^{2n+1}=121\cdot11^n+12\cdot144^n=12(144^n-11^n)+133\cdot11^n$$ and from here it's obvious because $$144^n-11^n=(144-11)\left(144^{n-1}+...+11^{n-1}\right)$$ is divisible by $133$.
For your new problem we have: $$11^{n+2}+12^{2n+1}=11\cdot11^{n+1}+144\cdot12^{2n-1}=144(11^{n+1}+12^{2n-1})-133\cdot11^{n+1},$$ which ends the proof.
The problem should be $11^{n+1}+12^{2n-1}$ is divisible by $133$.
Base step is trivial.
Inductive step: $$11^{k+2}+12^{2k+1}=11\times 11^{k+1}+144\times 12^{2k-1}$$ $$=11(11^{k+1}+12^{2k-1})+133\times 12^{2k-1}$$
I think the problem is wrong. The problem should be $11^{n+1}+12^{2n-1}$ is divisible by $133$.
