-1

Here is the question: Let $m=2^{15}−1=32767$. Show that

(a) The order of $2$ modulo $m$ is $15$.

(b) $15$ does not divide $m − 1$. Why does this imply that $m$ is not prime?

I have shown that the order of $2$ modulo $m$ is $15$, and that $15$ does not divide $m-1$. However, I am struggling with showing why this implies that $m$ is not prime. Can anybody provide any hints? I have tried to use Euler's Theorem to no avail.

Bill Dubuque
  • 272,048
Anon
  • 423
  • 2
  • 4
  • 1
    prime $m = 2^{15}!-1\overset{\rm Fermat}\Rightarrow \bmod m!:\ 2^{:!\large\color{#c00}{m-1}}!\equiv 1\Rightarrow {\rm ord_m}(2)!=!15\mid \color{#c00}{m!-!1},$ by Corollary$_1$ in the linked dupe. $\ \ $ – Bill Dubuque Sep 30 '23 at 03:54
  • For $~a,k \in \Bbb{Z^+},~$ you have that $~\left(x^a - 1\right)~$ divides $~\left(x^{ak} - 1\right).$ – user2661923 Sep 30 '23 at 05:04

1 Answers1

2

If $m$ were prime, then the multiplicative group $G=(\Bbb Z/m \Bbb Z)^{\ast}$ would have order $m-1$. You have shown that $2$ is an element of $G$ with order $15$, so $15 \mid \vert G \vert$. Since you also have shown that $15 \nmid (m-1)$, that means $\vert G \vert \neq m-1$, which in turn means $m$ can't be prime.

Robert Shore
  • 23,332