Prove this modular congruency? Prove that $2^{2x} \equiv 1 \mod 3$ for any integer $x$?. I know this is true but is there a nice way to prove it?

**HINT:-**

An approach which does not involve modular arithmetic.

R.T.P.

$2^{2x} \equiv 1 \mod 3$

or,$2^{2x}-1$ is divisible by $3$.

Now,$2^{2x}=4^x$

So,prove by induction now that $4^x-1$ is divisible by $3.$.'