<s>[INST] Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \cdot 1} = 9^2 + 7^2 = 130
$$
Therefore, $10|(9^{n+1}+7^{2n})$, is true.
Inductive Step: There exists an integer $k$ such that $(9^{n+1}+7^{2n}) = 10k$. Let $P(k) = 9^{k+1} + 7^{2k}$. Therefore,
$$
P(k+1) = 9^{(k+1)+1} + 7^{2(k+1)} = 9^{k+2} + 7^{2k+2}
$$
$$ 
P(k+1) = (9^1 \cdot 9^{k + 1}) + (7^2 \cdot 7^{2k})
$$
$$P(k+1) = (2+7)(9^{k+1}) + (7^2 \cdot 7^{2k}) = (2 \cdot 9^{k+1}) + (7 \cdot 9^{k+1}) + (7 \cdot 7 \cdot 7^{2k})
$$
$$
P(k+1) = (2 \cdot 9^{k+1}) + 7(9^{k+1} + 7 \cdot 7^{2k})
$$
 [/INST] By the binomial theorem, $$9^{n+1}+7^{2n}=(10-1)^{n+1}-(50-1)^n=(10a+(-1)^{n+1})-(50b+(-1)^n)=10a-50b$$ since $n$ and $n+1$ have different parities.
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