Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:

$4^n+5^n+6^n$ is divisible by 15 for positive odd integers

For $n=2k-1,n≥1$ (odd integer) 
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?

Hint :
\begin{eqnarray*}
31(4^n+5^n+6^n)= 4^{n+2}+5^{n+2}+6^{n+2} +15 \times 4^n+ 6 \times 5^n - 5 \times 6^n. 
\end{eqnarray*}
