<s>[INST] Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ 
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3.

In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So what I have done is said all primes $>3$ are odd $\implies n = 2k + 1$ for some $k \in \mathbb{N}$. Checking $2^1$ and $2^3$, we see that $2^{2k+1} \equiv 2 \mod 3$. From the lecture notes, I know that $2^{2k} \equiv 1 \mod 3$. I can then write
$$2^{2k + 1} = \underbrace{2^{2}\times 2^{2} \times \cdots \times 2^{2}}_\text{k times} \times 2 \equiv (1 \times 1 \times \cdots \times 1) \times 2 \equiv 2 \mod 3.$$
So we get that
$$2^{n} - 1 \equiv (2 \mod 3) - 1 \equiv 1 \mod 3$$
and so $3$ does not divide $2^n - 1$ when $n$ is odd. How does this show for it not dividing by $7$?
 [/INST] In fact we can prove more general result.
This holds true any number of the form $3k\pm1$  where $k$ is an integer
because $2^3=8\equiv1\pmod 7\implies 2^{3k}\equiv1$
$\implies 2^{3k+1}\equiv2\pmod 7\not\equiv1$
and $2^{3k-1}=2^{3(k-1)}\cdot2^2\equiv4\pmod 7\not\equiv1$
We know any prime $>3$ can be written as $6r\pm1$ (where $r$ is an integer) which is a proper subset of the numbers of the form $3k\pm1$
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