Show that $3^{4n+2} + 1$ is divisible by $10$

hint : $3^{4n+2} = (10-1)^{2n+1}$.

Hint: $$3^{4n+2}=3^{4n}\cdot 9$$ and $$3^{4n}\equiv 1 \pmod{10}$$ since $\phi(10)=4$.

What is the formation law of the remainders of the division by $10$ of the powers $3^{n}$?

(For the notation see modular arithmetic.)

$$\left\{ \begin{array}{c} 3\equiv 3\quad \pmod{10} \\ 3^{2}\equiv 9\quad \pmod{10} \\ 3^{3}\equiv 7\quad \pmod{10} \\ 3^{4}\equiv 1\quad \pmod{10} \end{array}\right. $$

$$\left\{ \begin{array}{c} 3^{5}\equiv 3\quad \pmod{10} \\ 3^{6}\equiv 9\quad \pmod{10} \\ \cdots \\ \cdots \end{array}\right. $$

So, with respect to the divisor $10$ the remainders of $3^{n}$ are periodic ($3,9,7,1,3,9,7,1,\dots$) with period $4$. This together with $4n+2\equiv 2\quad \pmod{4}$ yields $3^{4n+2}\equiv 3^{2}\quad \pmod{10}$. Also $1\equiv 1\quad \pmod{10}$. Hence, for all $n\ge 1$ $$3^{4n+2}+1\equiv 3^{2}+1\equiv 0\quad \pmod{10},$$

which means the remainders of the devision of $3^{4n+2}+1$ by $10$ are $0$.