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?

If $P(n)$ holds, then$$\begin{align}4^{2(n+1)+1}+3^{n+1+2}&=4^{2n+3}+3^{n+3}\\&=4^2\times4^{2n+1}+3\times3^{n+2}\\&=13\times4^{2n+1}+3\times\bigl(4^{2n+1}+3^{n+2}\bigr).\end{align}$$Since $13\mid4^{2n+1}+3^{n+2}$, $13$ divides this number.
