Is this formula true for $n\geq 1$:$4^n+2 \equiv 0 \mod 6 $? Is this formula true for $n\geq 1$:$$4^n+2 \equiv 0 \mod 6 $$.
Note :I have tried for some values of $n\geq 1$ i think it's true such that
:I used the sum digits of this number:$N=114$,$$1+1+4\equiv 0 \mod 6,1²+1²+4²\equiv 0 \mod 6,1^3+1^3+4^3\equiv 0 \mod 6,\cdots $$ ?
Thank you for any help 

For a less number theoretic approach, let $S_n$ be the statement that $4^n + 2$ is divisible by 6. Clearly $S_1$ is true since $4^1 + 2 = 6$. Now assuming that $4^k + 2$ is divisible by 6,
\begin{align*}
4^{k+1} + 2 &= 4\cdot4^n + 2 \\
&= 4\cdot4^n + 4\cdot2 - 6  \\
&= 4(4^n + 2) - 6,
\end{align*}
we immediately have that $4^{k+1} + 2$ is divisible by 6, so the induction is complete.
The other answers are far more elegant, however, and you should lean towards using modular arithmetic to solve these problems.
