This problem may be a lot harder than level 3.  N must be odd and 2^N is congruent to 1 mod N. Thus N is a multiple of e(N), the smallest of the positive integers x such that 2^x is congruent to 1 mod N. It is well-known that e(N) is a divisor of Euler's function phi(N) that gives the number of positive integers that do not exceed N and are coprime to N. However, it seems to require an algorithm to find which divisor of phi(N) equals e(N) . And while phi(N) itself cannot divide N, it is quite possible for a factor of phi(N) to divide N: for N=57 we have phi(N)=36, e(N)=18, and 3 divides both 57 and 18. Of course 18 does not divide 57.

While it does look like no odd N greater than 1 is divisible by e(N), the fact that the value of e(N) is so elusive appears to make this very tricky to prove.