$13\mid4^{2n+1}+3^{n+2}$ How can I prove that $4^{2n+1}+3^{n+2}$ is always divisible by 13?

You can use congruence:
$$
4^{2n+1}+3^{n+2}\equiv 4(16^n)+9(3^n) \equiv 4(16^n)+(13-4)(3^n) \equiv 4(16^n-3^n) \equiv 4(3^n-3^n) \equiv 0\ (\text{mod }13).
$$
Also you can prove it directly:
$$
4^{2n+1}+3^{n+2} = 4(16^n-3^n)+(13)(3^n) = 13[4(16^{n-1}+16^{n-2}3+16^{n-3}3^2+\ldots+3^{n-1})+3^n]
$$
which proves that $4^{2n+1}+3^{n+2}$ is dividable by $13$. 
