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?

By the binomial theorem,
$$
4^{2n+1} + 3^{n+2}
=4\cdot 16^n+9\cdot 3^n
=4\cdot (13+3)^n+9\cdot 3^n
=4(13a+3^n)+9\cdot 3^n
=13(4a+3^n)
$$
