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?

$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$

How do I substitute the statement where n=2k−1 to the above

By factoring one more power out...
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}=(16)4^{2k-1} + (25)5^{2k-1} + (36)5^{2k-1}$
So this is $[16(4^{2k-1} + 5^{2k-1}+6^{2k-1})] + 9*5^{2k-1} + 20*6^{2k-1}$.
And it's easy to finish:
$=[16*15N] + 3*15*5^{2k-2} + 4*15*2*6^{2k-2}$.
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But if you know modulo arithmetic this is CUTE!
$4^{n} + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n \equiv 1^n+(-1)^n + 0^n \equiv 0 \pmod 3$ so $3|4^n + 5^n +6^n$.
And $4^n + 5^n + 6^n = (5-1)^n + 5^n + (5+1)^n\equiv (-1)^n + 0^n + 1^n \equiv 0 \pmod 5$ so $5|4^n + 5^n +6^n$.
So $15|4^n + 5^n +6^n$.
....
If you don't know modulo arithmetic the you can use binomial theorem.
$4^n + 5^n + 6^n =(5-1)^n + 5^n + (5+1)^n =$
$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5 - 1) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5 + 1)=$
$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5)$
Which is divisible by $5$.
Do the same for $4^n + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n$ to show it is divisible by $3$.
