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 $10$.
 A: hint : $3^{4n+2} = (10-1)^{2n+1}$.
A: Hint: $$3^{4n+2}=3^{4n}\cdot 9$$ and $$3^{4n}\equiv 1 \pmod{10}$$ since $\phi(10)=4$.
A: 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 $0$.
A: $$9+1=10$$
$$81 \times \left( 3^{4n-2} +1 \right) = 3^{4n+2} + 1 +80$$
A: HINT $\rm\ \ \ A^K\: \equiv -1\ \ \Rightarrow\ \ A^{\:J+2\:K}\ \equiv\ A^{\:J}$
A: $3^{4n+2} = 3^{2(2n+1)} = 9^{2n+1}$.  So your question is really asking about odd powers of 9 and then adding 1.  Let's look at powers of 9, and focus on the last digit (ie, take them mod 10):
$9^1 = 9,\ 9^2 = 1,\ 9^3=9^1\cdot 9^2=9 \cdot 1 = 9, \cdots$.  So the last digit in consecutive powers of 9 goes 9, 1, 9, 1, 9, 1, ...
Now consider adding 1 to only the odd terms above.
A: $3^ {4*n} $ always gives a number ending in 1. Multiplying it by 9 gives a number ending in 9.
Adding 1 to it always gives a 0 at end.
Hence divisible by 10
