How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$  is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
Assume the statement is true for some integer $k\geq 2$.
$$f(k)=5^{2k+1}-3^{2k+1}-2=30t$$
For $n=k+2$,
\begin{align}
f(k+2)&=(5^{2(k+2)+1})-(3^{2(k+2)+1})-2\\
&=(5^4)\cdot (5^{(2k+1)})-(3^4)\cdot (3^{(2k+1)})-2
\\&=30(80t)+160+544\cdot (5^{2k+1})
\end{align}
I'm stuck here. Any help would be appreciated!

Since $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}-5-3\cdot3^{2n}+3=5^{2n+1}-5-3\left(81^{\frac{n}{2}}-1\right)=$$$$=5^{2n+1}-5-3\cdot80\left(81^{\frac{n}{2}-1}+...+1\right),$$
we see that $5^{2n+1}-3^{2n+1}-2$ is divisible by $5$.
Also, $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}+1-\left(3^{2n+1}+3\right)=(5+1)(5^{2n}-...+1)-\left(3^{2n+1}+3\right),$$
which says that $5^{2n+1}-3^{2n+1}-2$ is divisible by $6$.
Id est, $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$.
