Prove by induction that $4^n+5^n+6^n$ is divisible by $15$ for all odd $n\in \mathbb{N}$. 
Prove by induction that $4^n+5^n+6^n$ is divisible by $15$ for all odd $n\in \mathbb{N}$.

My proof :
I'm straight going to the induction hypothesis part.
Let for some odd $m(\gt2)\in \mathbb{N}, 15 | 4^m+5^m+6^m$.
Then, $$ 4^m+5^m+6^m = 15q$$ for some $q\in \mathbb{Z}$.
Now, $$4^m = 15q-5^m-6^m$$
Since $m$ is odd then the next odd number is $m+2$.
Now, $$ 4^{m+2} + 5^{m+2} + 6^{m+2} \\
= 4^m \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\
= (15q-5^m-6^m) \cdot 16 + 5^m \cdot 25 + 6^m \cdot 36 \\
= 15 \cdot 16q + 9 \cdot 5^m + 20 \cdot 6^m \\
= 15 \cdot 16q + 9 \cdot 5 \cdot 5^{m-1} + 20 \cdot 6 \cdot 6^{m-1} \\
= 15\big( 16q + 3 \cdot 5^{m-1} + 8 \cdot 6^{m-1} \big) \\
\\
\implies  15 | 4^{m+2} + 5^{m+2} + 6^{m+2} $$
Thus from induction the statement is true for every odd $n\in \mathbb{N}$
Is my proof correct?

Alternatively, let $f(n)=4^n+5^n+6^n$. Then
$$
f(n+2)-f(n)=15 \cdot 4^n + 24 \cdot 5^n + 35 \cdot 6^n = 15 \cdot 4^n + 15 \cdot 8 \cdot 5^{n-1} + 15 \cdot 14 \cdot 6^{n-1}
$$
So if $15$ divides $f(n)$, then $15$ divides $f(n+2)$. Note that $n+2$ is the next odd number if $n$ is odd.
