Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $
Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13  = \bigg( ( 4^{2n} \mod 13) \cdot ( 4 \mod 13 ) \bigg) \mod 13. $ But still I don't know how to simplify the first term in the large bracket.
Any help/suggestions?

We prove that $13$ divides $4^{2n+1}+3^{n+2}$ by induction on $n$.  If $n=1$, then $4^3+3^3=64+27=91$ is a multiple of 13.  So the assertion holds if $n=1$.  
Now assume that the assertion holds for $n=k$ (this assumption is called the inductive hypothesis).  Thus, $13$ divides $4^{2k+1}+3^{k+2}$. We prove that the assertion holds for $n=k+1$. We have that $4^{2n+1}+3^{n+2} = 4^{2(k+1)+1}+3^{(k+1)+2} = 16 \cdot 4^{2k+1} + 3 \cdot 3^{k+2}$.  In order to use the inductive hypothesis, we rewrite this last expression as $3\{4^{2k+1}+3^{k+2} \} + 13 \cdot \{ 4^{2k+1}\}$.  The first term is a multiple of 13 by the inductive hypothesis, and the second term is clearly a multiple of 13 because it has 13 as a coefficient, whence the sum is a multiple of 13.
