IMO 1964 - Problem 1

Problem: (a) Find all natural numbers $n$ such that the number $2^n-1$ is divisible by $7$.
(b) Prove that for all natural numbers $n$ the number $2^n+1$ is not divisible by $7$.
Solution: Let $n=3k+r$, where $0\le r<2$. Then $2^{n}=2^{3k+r}=8^{k}.2^{r}\equiv 2^{r}(\text{ mod }7)$.
Thus the remainder of $2^n$ modulo $7$ is $1,2,4$ if $n\equiv 0,1,2(\text{ mod }3)$. Hence $2^{n}-1$ is divisible by $7$ if and only if $3|n$, while $2^{n}+1$ is never divisible by $7$.