we have that 2^n \equiv 1 mod (2^n-1) therefore the residue class 2 has order n mod 2^n-1. I am intrested in finding a p(prime) such that 2 has order n in where p is prime. In case 2^n-1 is prime then I can apply the above argument and it is trivial to find such a p. But if 2^n-1 is compostie then I see that the largets prime in prime factorization of works (in cases where 2^n-1 has small primes in its factorization) For example M_4=2^4-1 =3*5 2 has order 4 in (Z/5)^* . and how can I be sure that there is always such a divisor. I hope my question is clear.