Problem
NT-B2-M03-P008 A Divisor of \(2^m-1\)
#8
★★★☆☆ Level 3 of 5
Let an odd prime \(p\mid2^m-1\). Prove that \( \operatorname{ord}_p(2)\mid \gcd(m,p-1) \).
The order divides both the exponent and \(p-1\).
From \(2^m\equiv1\pmod p\), we get \( \operatorname{ord}_p(2)\mid m \). Since \(p\) is prime and \(2\not\equiv0\pmod p\), the order also divides \(p-1\). Therefore it divides \(\gcd(m,p-1)\).
This problem is often the first step in restricting prime divisors.