<s>[INST] Prove by mathematical induction, that $81\times 3^{2n} - 2^{2n}$ is divisible by $5$ where $n \in \mathbb{Z}^+$ 
For all $k$, the equation I came up with is $3^{4+2k} = 5m + 2^{2k}$ where $m$ is a positive integer.
For all $k+1$, the expression is $3^{6+2k} - 2^{2k+2}$.

I tried to plug in the first equation to reach an expression that can be expressed with the integer $5$ taken common, but I am unable to figure out the required manipulation of the expressions to reach the result.
 [/INST] $$81\cdot3^{2n}-2^{2n}=5k\implies
\\\begin{align}81\cdot3^{2(n+1)}-2^{2(n+1)}=81\cdot9\cdot3^{2n}-4\cdot2^{2n}&=9\cdot(81\cdot3^{2n}-2^{2n})+5\cdot2^{2n}
\\&=9\cdot5k+5\cdot2^{2n}
\\&=5k'.\end{align}$$

Shorter:
$$81\cdot3^{2n}-2^{2n}=9^n-4^n\mod 5\implies 9^{n+1}-4^{n+1}=9\cdot9^n-4\cdot4^n=4(9^n-4^n)\mod5.$$

Yet shorter:
$$81\cdot3^{2n}-2^{2n}=9^n-4^n=4^n-4^n\mod 5.$$
</s>