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?

You could go mod $3$ and mod $5$ and conclude, an alternate proof is by induction : of course $4^n + 5^n + 6^n$ is divisible by $15$ when $n=1$. However, note that if $n \geq 3$:
$$
4^n + 5^n + 6^n - 4^{n-2} - 5^{n-2} - 6^{n-2} \\=\color{blue}{(4^n - 4^{n-2})} + \color{green}{(5^n - 5^{n-2})} + \color{red}{(6^n - 6^{n-2})} \\= \color{blue}{15(4^{n-2})} +\color{green}{ 15(8 \times 5^{n-3})} + \color{red}{ 15(14 \times 6^{n-3})}
$$
where terms of same colour are equal by factorization. Thus, the claim follows since the sum of two multiples of $15$ is also a multiple of $15$.
