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!

Suppose $f(k)=5^{2k+1}-3^{2k+1}-2=30t$
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}) \\
&= 30(80t) + 150 + 10 + 540(5^{2k+1}) + 4(5^{2k+1})\end{align}
It suffices to prove that $10+4(5^{2k+1})$ is divisible by $30$.
\begin{align}
10 + 4(5^{2k+1}) &= 10 + 20(5^{2k})
\end{align}
It is clearly divisible by $10$, we just have to show that it is divisible by $3$.
$$10+20(5^{2k}) \equiv 1-2^{2k}\equiv 1-4^k = 1-1^k \equiv 0 \pmod{3}$$
Hence is it divisible by $3$.
