Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$. Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. Most likely, due to the late hour, I am missing something very trivial here.

Prove that for any integer $n$, the number $11^{n+1}+12^{2n-1}$ is divisible by $133$.

I have tried multiplying through by $12$ and rearranging, but like I said with meager results. I arrived at $11^{n+2}+12^{2n+1}$ which satisfies the induction hypothesis for LHS, but for the RHS I got stuck at $12 \times 133m-11^{n+1}-11^{n+3}$ or $12^2 \times 133m-12 \times11^{n+1}-11^{n+3}$ and several other combinations none of which would let me factor out $133$. 

In fact, we can prove a stronger result and the proof is easier. The result we will prove is that $$x^2+x+1 \text{ divides }x^{n+1} + (x+1)^{2n-1}$$ for all $n \in \mathbb{N}$. Setting $x=11$ gives the result, you are looking for.
The proof follows immediately from the remainder theorem since $(x^2+x+1) = (x-\omega)(x-\omega^2)$, where $\omega$ is the complex cube-root of unity.
(Remember that $(x-a)$ divides $f(x)$ if and only if $f(x) = 0$)
Plugging in $\omega$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{n+1} + (\omega+1)^{2n-1} = \omega^{n+1} + (-\omega^2)^{2n-1} = \omega^{n+1} - \omega^{4n-2} = \omega^{n+1} - \omega^{3n} \omega^{n-2}\\
=\omega^{n+1} - 1 \times \omega^{n-2} = \omega^{n-2} \left( \omega^3 - 1\right) = 0$$
This gives us that $(x- \omega)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Similarly, plugging in $\omega^2$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{2n+2} + \left( \omega^2 + 1 \right)^{2n-1} = \omega^{2n+2} + (-\omega)^{2n-1} = \omega^{2n+2} - \omega^{2n-1}\\ = \omega^{2n-1} \left( \omega^3 - 1\right) = 0$$ This gives us that $(x- \omega^2)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Hence, $(x-\omega)(x-\omega^2) = x^2 + x + 1$ divides $x^{n+1} + (x+1)^{2n-1}$.
Setting $x=11$ gives us the result you want i.e. $11^2 + 11 + 1 = 133$ divides $11^{n+1} + 12^{2n-1}$.
Note that the above result can also be proved by inducting on $n$.
