Basic number theory proofs 
  
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*Deduce that there is a prime gap of length $\geq n$ for all $n \in \mathbb{N}$
  
*Show that if $2^n - 1$ is prime, then $n$ is prime.
  
*Show that if $n$ is prime, then $2^n - 1$ is not divisible by $7$ for any $n > 3$.
  



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*I'm not really sure how to do the first bit.

*For the second one, I'm not sure if this is correct but I have this: If $n$ is not prime, then $n = a b, a, b > 1$. Also $2^n - 1 = (2 - 1)(1 + 2 + 2^2 + .... + 2^{n-1})$. So if we substitute for $n = ab$, we get $(2^a)^b - 1 = (2^a - 1)(1 + 2^a + (2^a)^2 + .... + (2^a)^{b - 1})$ which is now composite and so if $n = ab$, then $2^n - 1$. This means that $2^n - 1$ is prime if $n$ is prime.

*For the last one, we have a hint "Follow the example in lectures to show that $2^n - 1$ is not divisible by $3$", but I missed this lecture and haven't had a chance to get the note, so can someone help me with the divisible by $3$ bit and then I can try the divisible by $7$ bit myself.
Thank you
EDIT: I found a simpler proof for the second bit online:

Supposed $n$ is composite, it can be then written in the form $n = ab, a, b > 1 \implies (2^a)^b - 1$ is prime with $b>1$ and $2^a>2$, contradicting statement $1$.

What is statement $1$? Maybe then I can understand why it contradicts it.
 A: Since you have shown that there are no primes between $n!$ and $n!+n$, now it's time to combine this with the infinitude of prime numbers and we have the following:
Claim: Let $n>2$ be a natural number, let $p$ be the maximal prime number such that $p<n!$ and let $q$ be the least prime number such that $n!<q$, then there is no prime number between $p$ and $q$ and $q-p\geq n$.
Proof. The first part of the claim is true because any number between $p$ and $q$ would either be between $p$ and $n!$ in contradiction to how we chose $p$, or between $n!$ and $q$ which is a contradiction to how we chose $q$. The second part follows from the fact that $n!+i$ is not a prime number for any $i<n$ and therefore $p<n!+i<q$ and so $q-p\geq n$.
A: For the first part, you can use the idea of a factorial. Since you are looking for, say $n$ gap, start with $n!+2$. Then $n!+2$ can be divided by 2, since $n!=n(n-1)...2.1$ . Also $n!+k$ if $k$ is less than $n$ divides by $k$, and obviously $n!+n$ divides by $n$. That's a gap of $(n-1)$. If you want $n$ gap, start with (n+1)!.  
A: For the second, you start with "Assume $n$ is not prime, (in which case you want to consider separately the case $n = 1$ which is not prime, and the case $n$ is composite"), so...(there exist $a>1$, $b>1$ such that $n = ab$).
Then you want to end with "hence, if $n = ab$, then $2^n - 1$ cannot be prime. So you've proven the contrapositive of the proposition, and should end with confirming the proposition: "if $2^n - 1$ is prime, then $n$ must be prime."
Regarding your edit, I am assuming the statement 1 was an assumption, for the sake of contradiction, $n$ is not prime. So $n = 1$ or $n = ab, a, b>1$, and showing case 1: if $n = 1$, then $2^n - 1 = 2 - 1 = 1$, which is not prime. Then what you see is "case 2"... and conclusion.
A: Hint: For the second part, argue by contrapositive as amWhy suggested. Suppose that $n = ab$. Show that $2^n - 1 = 2^{ab} - 1$ is composite by considering the difference of powers factorization
$$x^k - y^k = (x-y)(x^{k-1} + x^{k-2}y + \cdots + xy^{k-2} + y^{k-1})$$
It'll also be instructive to reason out why the method fails for prime exponent.
For the third part, consider 
$$2^n - 1 \pmod 7$$
As $n$ ranges through the natural numbers, the congruence repeats in a very obvious pattern. Try to reason out exactly when $2^n \equiv 1 \pmod 7$.  
A: For the 3rd part, if $7\mid(2^n-1)$ where  $n$ is any natural number,
We get $2^n\equiv1\pmod 7$
Now, $2\equiv2\pmod 7,2^2\equiv4,2^3=8\equiv1\pmod 7\implies ord_72=3$
Form here,  $ord_72\mid n\implies 3\mid n$
Now if $n$ is prime, $n$ must be $3$ else $n$ will be composite.
