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 
 A: The answer is yes, because $4^2\equiv 4\pmod{6}$, and hence $4^n\equiv 4\pmod{6}$ for all $n\ge 1$.
A: For all $n$,
$$
4^n\equiv1^n\equiv1\pmod{3}
$$
For all $n\gt0$,
$$
4^n\equiv0^n\equiv0\pmod{2}
$$
Therefore, by the Chinese Remainder Theorem
$$
4^n\equiv4\pmod{6}
$$
and therefore, for all $n\gt0,$
$$
4^n+2\equiv0\pmod{6}
$$
A: 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.
A: Let us make $\pmod2$ and $\pmod 3$ separately.
It very simple.
$2$ divides $4^n+2$ so for the first step we get $4^n+2\equiv0\pmod2$ Obviously.
As $4\equiv1\mod3$, $4^n\equiv1\pmod3 \space \forall n>0$, and therefore $4^n+2\equiv0\pmod3$
$4^n+2\equiv0\pmod2$ and $4^n+2\equiv0\pmod3$
It follows that $4^n+2\equiv0\pmod6$
Q.E.D.
