Does $3$ divide $2^{2n}-1$? Prove or find a counter example of : $3$ divide $2^{2n}-1$ for all $n\in\mathbb N$.
I compute $2^{2n}-1$ until $n=5$ and it looks to work, but how can I prove this ? I tried by contradiction, but impossible to conclude. Any idea ?

HINT: $$2^{2n}-1=4^n-1\equiv 1^n-1=1-1\equiv 0\mod 3$$
