<s>[INST] 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?
 [/INST] Try induction. For $n=0$ the term $4^{2n+1} + 3^{n+2}$ becomes $4 + 9 = 13$, so that's OK.
Then suppose 13 divides $4^{2n+1} + 3^{n+2}$, then consider $4^{2(n+1)+1} + 3^{(n+1)+2} = 4^{2n+3} + 3^{n+3}$. 
This equals $16 \cdot 4^{2n+1} + 3 \cdot 3^{n+2} = 13 \cdot 4^{2n+1} + 3\cdot( 4^{2n+1} + 3^{n+2})$, which is divisible by $13$ if the number in brackets is.
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