Find all positive integers n for which n divides 2^n-1.
Only possible values of n is 1
Proof:
Let p \mid n be the smallest prime divisor of it.
and let \text{ord}_p(2) = r then we have:
p \mid 2^{n}-1 \Rightarrow r \mid n But we know: r \mid p-1. Hence we conclude that r \mid \text{gcd}(n,p-1) = 1 because p is the smallest prime.
Hence r=1 \Rightarrow p=1 contradiction.