How do I show that for any natural number n, there exists a natural number m such that $4^{2n+1} + 3^{n+2} = 13m$? 
Show that for any natural number $n$, there exists a natural number $m$ for which:
  $$4^{2n+1} + 3^{n+2} = 13m$$

I don't know where to start. I tried to use Mathematical Induction, denoting the top statement by $\rm P(n)$ and prove that $\rm P(0)$ is true, but I got stuck. Can anybody help?

The statement $P(n)$ is:

There exists $m\in\mathbb N$ such that $$4^{2n+1}+3^{n+2}=13m$$


Step $1$: Proving that $P(0)$ is true should be easy, you just have to calculate what $$4^{2\cdot 0+1}+3^{0+2}$$ is equal to.

Step $2$: Assume that $P(n)$ is true, and write
$$\begin{align}4^{2(n+1)+1}+3^{(n+1)+2} &= 4^{2n+1+2} + 3^{n+2 + 1} \\&= 16\cdot 4^{2n+1} + 3\cdot 3^{n+2}\\&=13\cdot 4^{2n+1} + 3\cdot 4^{2n+1} + 3\cdot 3^{n+2}\end{align}$$
Can you continue from here?
