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!
 A: 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 $n$.
A: Here is an intuitive way to do an inductive proof that doesn't require knowledge of modular arithmetic yet remains faithful to its arithmetical genesis.
Take notice that $\ 5\cdot 25^n - 3\cdot 9^m - 2\,$ is a special case of below
Theorem $\,\ 30\mid 5(1\!+\!6a)^n - 3(1\!+\!10b)^m - 2,\,$ for $\,a,b\in\Bbb Z,\, n,m\in\Bbb N.\ \ \ $ Proof:
By induction on $\,n\,$we can show $\, 5(1+6a)^n\! =\, 5 + 30j\ $ for some $\ j\in\Bbb Z.\,$ By the same proof
by induction on $m$ we can show $\,3(1\!+\!10b)^m\! =\:\! 3 + 30k\,$ for some $\,k\in \Bbb Z$
Subtracting them and $\,2\,$ yields $\,5(1\!+\!6a)^n - 3(1\!+\!10b)^m - 2 = 30(j-k).\ \ {\small\bf QED}$
Remark $ $ Here the employed induction is much more intuitive, essentially amounting to an inductive proof of the first couple terms of the Binomial Theorem, as here. When you learn modular arithmetic such inductions become neatly encapsulated for handy reuse by applications of the Congruence Power Rule, as explained here. In particular we can use the handy mDL law
$$\begin{align} ab\bmod ac\, &=\, a(b\bmod c) = \text{$\!\bmod\!$ Distributive Law [mDL]}\\[.3em]
\Rightarrow\ 5(1\!+\!6a)^n \bmod 30\, &=\, 5((1\!+\!6a)^n\bmod 6) = 5(\color{#c00}{1^n}\!\bmod 6 ) = 5(\color{#c00}1)\end{align}\qquad\qquad\qquad$$
which reduces the proof to an extremely trivial induction $\,\color{#c00}{1^n\equiv 1}.\,$
A: 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 $3$.
A: 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 $30$.
A: 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 $960=\gcd(g(0),g(1),g(2))$.
