Proof Verification: Prove $15^n+6^{2n+1}$ is divisible by 7 for all $n\in \mathbb{N}$

Question

Let $P\left(n\right)$ be the statement that $15^n+6^{2n+1}=7m$ for some $m\in \mathbb{N}$

Prove the base case of $n=0$: $P\left(n\right)=15^0+6^1=7$ which is divisble by 7. Therefore, is true for $n=0$.

Assume $P(n)$ holds truth for some non-negative integer, $k$ $\Longrightarrow$ $P(k)=P\left(k\right)=15^k+6^{2k+1}=7m$ where $m\in \mathbb{N}$

We will now prove $P(k+1)$:

$P\left(k+1\right)=15^{k+1}+6^{2k+3}=15\left(15^k\right)+36\left(6^{2k+1}\right)$

$=15\left(7m-6^{2k+1}\right)+36\left(6^{2k+1}\right)$ from assumption

$=105m+21\left(6^{2k+1}\right)$

$=7\left(15m+3\left(6^{2k+1}\right)\right)$

Therefore, by the principle of mathematical induction, $P(n)$ is true for all $n\in \mathbb{N}$

Answer 1

$15^n + 6^{2n+1} = 1^n + (-1)^{2n+1} (mod7) = 1 - 1 = 0$ Another proof of your result

Answer 2

Your solution looks perfect.

Here is an alternative solution.
Note that, in congruence notation $15\equiv1\mod(7)$ and $6\equiv-1\mod(7).$ Hence $$15^m+6^{2n+1}\equiv1^m+(-1)^{2n+1}\mod(7)\equiv0\mod(7)$$ for any $n, m\in\mathbb{N}.$

Answer 3

$15^n+6^{2n-1}=15^n-1+6^{2m-1}+1=(15-1)L+(6+1)M=7(2L+M).$