divisibility - Prove that $4^{2n+1}+3^{n+2} : forall ninmathbb{N}$ is a multiple of $13$

How to prove that $\forall n\in\mathbb{N},\exists k\in\mathbb{Z}:4^{2n+1}+3^{n+2}=13\cdot k$

I've tried to do it by induction. For $n=0$ it's trivial.

Now for the general case, I decided to throw the idea of working with $13\cdot k$ and try to prove it via congruences. So I'd need to prove that $4^{2n+1}+3^{n+2}\equiv0\pmod{13} \longrightarrow 4^{2n+3}+3^{n+3}\equiv0\pmod{13}$, that is, $4^{2n+1}+3^{n+2}\equiv4^{2n+3}+3^{n+3}\pmod{13}$

But I have no clue how to do is. Any help?

Answer

$$4^{2n+3}+3^{n+3}= 16 \times 4^{2n+1} + 3 \times 3^{n+2} =13 \times 4^{2n+1} +3 \times(4^{2n+1}+3^{n+2})$$