<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] This is not an induction proof, but I'll nevertheless leave it here.
To show that $5^{2n+1}-3^{2n+1}-2$ is divisble by $30=2\cdot 3\cdot 5$, you can show that $2,3,5$ divides it.

*

*$2\mid 5^{2n+1}-3^{2n+1}-2\iff 2\mid 5^{2n+1}-3^{2n+1}$, but this is clear since the sum of two odd numbers is even.


*$3\mid 5^{2n+1}-3^{2n+1}-2\iff 3\mid 5^{2n+1}-2$, write $5^{2n+1}-2$ as $$(6-1)^{2n+1}-2$$ Upon expanding, every term is a multiple of $6$, and the last term is $(-1)^{2n+1}-2=-3$. So everything is divisible by $3$.


*$5\mid 5^{2n+1}-3^{2n+1}-2\iff 5\mid -3^{2n+1}-2$. Same trick, rewrite $-3^{2n+1}-2$ as $$-3(10-1)^n-2$$ Upon expanding every term except the last term is divisible by $5$, the last term is $-3(-1)^n-2$, so the result is only true for even $n$.
The claim is only true for even $n$.
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