1. ## 53|2^k-2^n

Find all $n,k\in\mathbb{N}$ for which $53|2^k-2^n$.

2. Originally Posted by james_bond
Find all $n,k\in\mathbb{N}$ for which $53|2^k-2^n$.

Choose $k>n\Longrightarrow k=n+m\,,\,m>0\Longrightarrow 53\mid (2^{n+m}-2^n)= 2^n(2^m-1)\iff$

$\iff 53\mid (2^m-1)\iff 2^m=1\!\!\pmod{53}$ , and since 2 is not a primitive

root modulo 53 (look here Primitive root modulo n - Wikipedia, the free encyclopedia) ,

the above is true only when...

Tonio

Edit: Either the above link is wrong or else I didn't understand it. Anyway, 2 indeed is

a primitive root modulo 53, and thus $2^m=1\iff m=k-n=0\!\!\pmod {52}$

3. Originally Posted by james_bond
Find all $n,k\in\mathbb{N}$ for which $53|2^k-2^n$.
We have $k\geq n$. Write $2^k-2^n=2^n(2^{k-n}-1)$. Since $(53,2^n)=1$, $53$ must divide $2^{k-n}-1$, i.e $2^{k-n}\equiv1\pmod{53}$.

Actually $2$ is a primitive root of $53$; hence the congruence holds if and only if $k-n\equiv0\pmod{52}$.

Now I think this is the answer:
For any fixed integer $t\geq0$, the solutions are given by any pair $k, n$ such that $k-n=52t$.