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?
HINT:
$$4^2\equiv3\pmod{13}\implies4^{2n}=(4^2)^n\equiv3^n$$
Now take out $3^n$ as common factor.
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.
It can be easily done by mathematical induction . If we put $n=0$,$1$ etc then it is true . Let $4^{2n+1}+3^{n+2}\equiv 0\pmod {13}$. We have to prove that $4^{2n+3}+3^{n+3}\equiv 0\pmod {13}$
Now:
$4^{2n+3}+3^{n+3}$.
=$16*4^{2n+1}+3*3^{n+2}$
=$13*4^{2n+1}+3*(4^{2n+1}+3^{n+2})$
=$13*4^{2n+1}+3*13X$
=$13*Y\equiv 0\pmod {13}$ .
By the binomial theorem, $$ 4^{2n+1} + 3^{n+2} =4\cdot 16^n+9\cdot 3^n =4\cdot (13+3)^n+9\cdot 3^n =4(13a+3^n)+9\cdot 3^n =13(4a+3^n) $$
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.
