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 ?

You can use the binomial theorem:
$$2^{2n}-1 = 4^n -1 = (3+1)^n -1 $$
$$= \left(3^n + {n \choose 1}3^{n-1} + {n\choose 2} 3^{n-2} + \cdots +{n\choose n-1}3 + 1 \right)-1.$$
When you cancel the last $-1$ with the $+1$, everything left has a factor of $3.$
Or, if you want induction, note that 
$$2^{2(n+1)}-1 = 4(2^{2n}) -1 = 4(2^{2n}-1) +4 -1 = 4(\mbox{multiple of 3}) +3. $$
