0
$\begingroup$

Title says it all. I was wondering if there was a clean way to do this, or if there are any conditions on $m$ if this is true.

  • 0
    So you are looking for conditions on $m$ given that $n$ is the smallest natural such that $$2^n \equiv 1 \pmod m$$ My knowledge of the multiplicative order function is kind of fuzzy. – 2017-01-13
  • 1
    $m$ has to be a divisor of $2^n-1$ and must not be a divisor of $2^{n/p}-1$ for any prime $p\mid n$. Whether this is useful depends on how far you can factor numbers of the form $2^\ell-1$. – 2017-01-13

1 Answers 1

1

First be sure that $\gcd(2,m)=1$ since both must be coprime in order to satisfy $2^n \equiv 1 \pmod m$. Then $m$ has to be odd. Also $n > \log_2(m)$

Note that $ord_m(2) = n \iff 2^n \equiv 1 \pmod m$ and as consequence of this we know that $n | \varphi(m)$.

A possible solution involving modulo a prime number:

If given $n\in\mathbb{N}$ ($n$ is even) and $n+1\in\mathbb{P}$ then $m=n+1$. Since $m$ is prime we will have $2^n \equiv 1 \pmod m$ by Fermat's Little.

Then $2^{n\cdot k} \equiv 1 \pmod m$ for $k\in\mathbb{N^+}$. This means that if $m=n+1$ is prime then $m$ is a solution for $(n,2n,3n....)$. $m$ is solution for an infinite set of multiples of n.