I am asked to prove that the product of all the elements in $ \mathbb{Z}_{m}^{*}$ for $ m>2$ equals to $ \pm {1} \mod{m} $. So far I tried the following direction. I defined: $$ S_{1} = \{x \in Z^*_{m} : x^2\equiv 1\mod{m}\} $$ and $$ S_{2} = \{x \in Z^*_{m} : x^2\not \equiv 1\mod{m}\} $$ If $P_{i}$ is the product of all the elements in $S_{i}$ for $i=1,2$, I proved that it holds $$ P = P_1\cdot P_2 $$ I already proved that $P_2=1 \mod m$. I need to prove $P_{1}= \pm 1 \mod{m} $. I already know that $P_{1}^2 = 1\mod{m} $ but unfortunately this doesn't imply the required. Can I get a hint on how to prove that $P_{1}= \pm 1 \mod{m} $? Thank you

  • $\begingroup$ A good start. One way of dealing with $S_1$ is to first consider the case when $m$ is a prime power (when that prime is equal to two there is an extra obstacle). Then apply the Chinese Remainder theorem to deduce $P_1$ modulo all the prime powers that appear in the factorization of $m$. Possibly there is an easier way forward. I had to wake up very early this morning, so I'm not firing on all cylinders at the moment. $\endgroup$ Apr 13 '18 at 20:10
  • 2
    $\begingroup$ Call $x,y \in S_1$ equivalent if $x = y$ or $x = -y$. Partition $S_1$ into equivalence classes and look at each separately first. $\endgroup$ Apr 13 '18 at 20:12
  • 1
    $\begingroup$ @DanielFischer This is brilliant. If $ x \in S_{1} $ then $ m-x \in S_{1} $ and $ x\cdot (p-x) \equiv -1 \mod{m} $. Thank you. $\endgroup$
    – Bes Dollma
    Apr 13 '18 at 20:21
  • $\begingroup$ @JyrkiLahtonen Thank you too $\endgroup$
    – Bes Dollma
    Apr 13 '18 at 20:21
  • $\begingroup$ @Ben See this answer for more on the method hinted by Daniel. $\endgroup$ Nov 23 '20 at 13:16

On $S_1$, introduce the equivalence relation

$$x \sim y \iff (x = y \lor x = -y)\,.$$

Then every equivalence class contains exactly two elements of $\mathbb{Z}_m^{\ast}$ (if $m > 2$ is even, then $m/2$ is not coprime to $m$, so we have $x \neq -x$ for all $x \in \mathbb{Z}_m^{\ast}$), and

$$x\cdot (m-x) = mx - x^2 \equiv -x^2 \equiv -1 \pmod{m}\,.$$


$$P_1 \equiv (-1)^{(\operatorname{card} S_1)/2} \pmod{m}\,.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.