<s>[INST] 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!
 [/INST] Write $g(n)=f(2n)=5\cdot 625^n-3\cdot 81^n-2\cdot1^n$. Then
$$
g(n+3)= 707 g(n+2) - 51331 g(n+1) + 50625 g(n)
$$
because $(x-625)(x-81)(x-1)=0$ can be written $x^3=707 x^2 - 51331 x + 50625$. (The important point here is that the coefficients are integers, not their exact values.)
Therefore, the claim follows at once by induction if it holds for $n=0,1,2$:
$$
g(0)=0, \quad g(1)=2880 = 30 \cdot 96, \quad g(2)= 1933440 = 30 \cdot 64448
$$
(No need to compute the quotient: it's easy to see that they are all multiples of $3$ and $10$.)
Actually, $g(n)$ is divisible by $960=\gcd(g(0),g(1),g(2))$.
</s>