2022/046) Find all n such that $7 | 2^n-1 $ and show that there is no positive n such that $7 | 2^n + 1$ (IMO 1964 problem 1)

because 7 is a prime so as per Fermats Little theoren 

$ 7 | 2^6-1$

now as $2^6$ leaves remainder 1 after dividing by 7 so it may be taht for some factor a of 6 $7 | 2^a-1$

we need to check for 1,2,3

so see $2^1-1 = 1$ $2^2-1 = 3$ and $2^3 - 1 = 7$ out f these 3 3 satisfies

as so k = 3m for all m satisfies.

Further $2^1+1 = 3$ $2^2+1 = 5$ and $2^3 +1 = 9$ out f these 3 none is divsible by 7 so there is no n such that $ 7 | 2^n+1$ 

proved