Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$ Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$
Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that
$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$
Thus $5 \mid 3^{4(k+1) + 2} + 1$.
Is this a valid proof? I'm not entirely sure I'm correct with this...
Hint: $3^4 \cdot 3^{4k+2} + 1= 3^4(3^{4k+2}+1)+1-3^4=3^4(3^{4k+2}+1)-80.$
An alternative strategy for this kind of problem is to consider the difference between consecutive terms.
Let $f(n)=3^{4n+2}+1$. Then $f(n+1)-f(n)=80 \cdot 3^{4 n + 2}$ is a multiple of $5$.
The claim follows by induction since $f(0)=10$ is a multiple of $5$.
(Actually, this proves that  $f(n)$ is always a multiple of $10$.)