<s>[INST] Prove that $7 | (3^{2n + 1} + 2^{n +2})$ Prove that $7 | (3^{2n + 1} + 2^{n +2})$
So far I have:
Base case:  n = 1
$ = (3^{2(1) + 1} + 2^{(1) +2})$
$ = (3^{3} + 2^{3})$
$ = (35)$ which divides 7
Inductive Step:
$ = (3^{2(n +1) + 1} + 2^{(n +1) +2})$
$ = (3^{2n + 3} + 2^{n + 3})$
$ 7j = (3^{2n + 3} + 2^{n + 3})$
-I'm a bit stuck as to how to proceed, any help or correcitons are appreciated.
- I believe I understand where we need to go now but i'm not sure how to get there. It seems obvious that any n plugged into this formula would be divisible by 7 given that $3^1 + 2^2 = 7$ Since you have that $3^1+2^2$ nested inside the given formula it seems easily understood that any sum with any positive n would also be divisible by 7. I just don't see the steps in between to actually prove it.
 [/INST] Let $7|3^{2k+1}+2^{k+2}$.
To prove that is divisible by $7|3^{2k+3}+2^{k+3}$, it suffices to prove that the difference is divided by
$$\left((3^{2k+3}+2^{k+3})-(3^{2k+1}+2^{k+2}) \right)=\left((3^{2k+1+2}+2^{k+2+1})-(3^{2k+1}+2^{k+2}) \right)=$$
$$=\left(9 \cdot 3^{2k+1}+2\cdot 2^{k+2}-3^{2k+1}-2^{k+2} \right)=$$
$$=8 \cdot 3^{2k+1}+2^{k+2}=7 \cdot 3^{2k+1} +\left(3^{2k+1}+2^{k+2} \right)$$
Obviously, both terms are divisible by 7, the second induction hypothesis
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