Proving $3^n + 4^n + 5^n$ is divisible by $12$ for all positive odd integers $n$ by induction I am trying to prove by induction that $\sum_{i=3}^{5} i^n$ is divisible by $12$ for all positive odd integers $n$. I heard that the solution might involve normal induction, where you assume that it works for the lowest positive odd number (in this case, $1$) and then try to deduce $P(n+2)$ for the next odd number.
I am not sure how to go about this, so any help would be greatly appreciated. I was thinking that if we define $n$ to be odd, then we could try to show that $n+2$ works using the fact that $n+2$ is also odd. However, I am not sure if this is the correct approach or how to proceed.
Could someone please help me with this problem?
Thank you!

We set $n = 2k - 1$ and carry an induction over $k$. For the base case, $k = 1$, $n = 1$ and $3 + 4 + 5 = 12$, which is clearly divisible by 12.  Assume true for $k$, and consider the case for $k + 1$. We have.
\begin{align*}
3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1} & = 9\cdot 3^{2k-1} + 16\cdot 4^{2k-1} + 25\cdot 5^{2k-1}\\
& = 8\cdot 3^{2k-1} + 15\cdot 4^{2k-1} + 24\cdot 5^{2k-1} + (3^{2k-1} +  4^{2k-1} + 5^{2k-1}).
\end{align*}
From our induction hypothesis, we know $3^{2k-1} +  4^{2k-1} + 5^{2k-1}  \equiv 0 \ \ \ (\text{ mod } 12 \ )$. Also, $8\cdot 3^{2k-1}$, $15\cdot 4^{2k-1}$ and $24\cdot 5^{2k-1}$ are all $\equiv 0 \ \ \ (\text{ mod } 12 \ )$ implying $3^{2k + 1} + 4^{2k + 1} + 5^{2k + 1}$ is divisible by 12.
