Show that $3^{4n+2} + 1$ is divisible by $10$ I'm am a little bit stuck on this question, any help is appreciated.
Show that for every $n\in\mathbb{N}$, $3^{4n+2} + 1$ is divisible by $10$.

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 $0$.
