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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

Bill Dubuque
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Bes Dollma
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  • 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. – Jyrki Lahtonen Apr 13 '18 at 20:10
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    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. – Daniel Fischer Apr 13 '18 at 20:12
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    @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. – Bes Dollma Apr 13 '18 at 20:21
  • @JyrkiLahtonen Thank you too – Bes Dollma Apr 13 '18 at 20:21
  • @Ben See this answer for more on the method hinted by Daniel. – Bill Dubuque Nov 23 '20 at 13:16

1 Answers1

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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

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

Hence

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

Daniel Fischer
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