Is this formula true for $n\geq 1$:$4^n+2 \equiv 0 \mod 6 $?

Question

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

Answer 1

The answer is yes, because $4^2\equiv 4\pmod{6}$, and hence $4^n\equiv 4\pmod{6}$ for all $n\ge 1$.

Answer 2

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 $n\gt0,$ $$ 4^n+2\equiv0\pmod{6} $$

Answer 3

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.

Answer 4

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.