Does $3$ divide $2^{2n}-1$?

Question

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 ?

Answer 1

A direct proof: $$2^{2n}-1=(2^n)^2-1=(2^n+1)(2^n-1).$$ Since $2^n$ is not divisible by 3, then either $(2^n+1)$ or $(2^n-1)$ must be divisible by 3.

Answer 2

Since $2^{2n}=4^n$, you get $$2^{2n}\equiv 1\pmod 3\implies 2^{2n}-1\equiv 0\pmod 3,$$ what prove the claim.

Answer 3

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

Answer 4

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

$$2^{2(n+1)}-1 = 4(2^{2n}) -1 = 4(2^{2n}-1) +4 -1 = 4(\mbox{multiple of 3}) +3. $$

Answer 5

Induction proof. Assume $2^{2n}-1$ is divisible by $3$.

Then $$\begin{align}2^{2(n+1)}-1&=2^2\cdot 2^{2n}-1\\ &=4\cdot (2^{2n}-1)+4-1\\ &=4(2^{2n}-1)-3 \end{align}$$

Since $2^{2n}-1$ is divisible by $3$, we get that $2^{2(n+1)}-1$ is divisible by $3$.

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If you know linear recurrences, you can see that if $$a_n=2^{2n}-1$$

then you can show that $$a_{n+1}=5a_{n}-4a_{n-1}$$

So if you show that $a_0$ and $a_1$ are divisible by $3$, then all $a_n$ are divisible by $3$.