1. (CZS) (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 .
1. Let $n=3 k+r$, where $0 \leq r<2$. Then $2^{n}=2^{2 k+r}=8^{k} \cdot 2^{r} \equiv 2^{r}(\bmod 7)$. Thus the remainder of $2^{-}$modulo 7 is $1,2,4$ if $n \equiv 0,1,2(\bmod 3)$. Hence $2^{n}-1$ is divisible by 7 if and only if $3 \mid n$, while $2^{n}+1$ is never divisible by 7 .