<s>[INST] 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?
 [/INST] Hint
Like Prove that $3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all values of $n$
If $f(m)=4^{2m+1}+5^{2m+1}+6^{2m+1},$
$$f(n+1)-4^2f(n)=5^{2n+1}(5^2-4^2)+6^{2n+1}(6^2-4^2)$$ will be clearly divisible by $15$ if $n\ge0$
So, if $15$ divides $f(n),15$ will divide $f(n+1)$
Now establish the base case i.e., $m=0$
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