Proof that $2^{222}-1$ is divisible by 3

How can I prove that $2^{222}-1$ is divisible by three? I already have decomposed the following one: $(2^{111}-1)(2^{111}+1)$ and I understand I should just prove that $(2^{111}-1)$ is divisible by three or that $(2^{111}+1)$ is divisible by three. But how can I solve this problem?

Solution 1:

The three numbers $2^{111}-1$, $2^{111}$ and $2^{111}+1$ are consecutive, and so one of them is divisible by $3$. But $2^{111}$ is not, since $2$ is prime.

Solution 2:

HINT

If you're familiar with binomial theorem $$\color{blue}{2^{222}}-1 = \color{blue}{(3-1)^{222}}-1 = \color{blue}{1+3M} -1 = \color{blue}{3M}$$

Solution 3:

The routine way is to invoke Fermat's little theorem: $$a^{p-1}-1\equiv 0\,(\text{mod}\,p)$$ for $\mathrm{gcd}(a,p)=1$. Plug in $a=2^{111},p=3$.

Solution 4:

When $n$ is odd we have that $$a^n+1=(a+1)(a^{n-1}-a^{n-2}+a^{n-3}-\cdots-a+1)$$ Plugging in $a=2$ you see that the expression is divisible by 3