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$x$ is a square in $(\mathbb{Z} /p \mathbb{Z})^\times$ iff there is a $y \in (\mathbb{Z} /p \mathbb{Z})^\times$ such that $x \equiv y^2 \mod p$.

I am asked to show that there are exactly $\frac{p-1}{2}$ squares in $(\mathbb{Z} /p \mathbb{Z})^\times$. How do I tackle this?

Praveen
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Marc
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1 Answers1

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You have of course to assume $p$ odd. Observe that the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^*$ has $p-1$ elements.

Consider the group homomomorphism $\rho\colon (\mathbb{Z}/p\mathbb{Z})^*\to (\mathbb{Z}/p\mathbb{Z})^*$ given by $x\mapsto x^2$. The image is the subgroup $S\subset (\mathbb{Z}/p\mathbb{Z})^*$ of squares. Hence, $S$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^*/\mathrm{Ker}(\rho)$. It remains to show that $\mathrm{Ker}(\rho)$ is of order $2$.

This can be for example seen by the fact that $(\mathbb{Z}/p\mathbb{Z})^*$ is cyclic.

Jérémy Blanc
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  • Nice, I hadn't thought of it like that before. – Asinomás Jan 07 '15 at 05:09
  • Thanks! I am missing a bit about the basics though. Why does this group have $p-1$ elements? And how does the cyclicity make the kernel of order two? – Marc Jan 07 '15 at 05:39
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    An element of order two would be contained in a subgroup of order $2$, cyclic subgroups have exactly one subgroup of each order. Alternatively suppose $x^2=1$, then $x^2-1$ is a multiple of $p$, By euclid's lemma $p$ divides $x+1$ or $p$ divides $x-1$. – Asinomás Jan 07 '15 at 06:11
  • Also note that $x\mapsto x^2$ is a homomorphism if and only if the group is abelian, but in this case it is so no problem. – Asinomás Jan 07 '15 at 06:12
  • @Marc, the group has $p-1$ elements: $1,2,\dots,p-1$. Indeed, $0$ is certainly not invertible but the other elements are: if $a\in \mathbb{Z}$ is not a multiple of $p$, it is coprime with it, so there exist $b,c\in \mathbb{Z}$ such that $ab+pc=1$. Hence, the class of $a$ in $\mathbb{Z}/p\mathbb{Z}$ is invertible. In order to show that the kernel has order $2$, observe that $\pm 1$ lie in the kernel, are distinct (because $p>2$) and that no other element lie in the kernel. This last observation was explained by Jorge Fernandez above, using the cyclicity or not. – Jérémy Blanc Jan 07 '15 at 07:41
  • Why is $(\mathbb{Z}/p\mathbb{Z})^\times$ cyclic? – Junglemath Jun 26 '20 at 14:26
  • @Junglemath: there are many ways to prove it. In general, every finite subgroup of $(K^*,\cdot)$ is cyclic, for each field $K$. For finite fields, there are also simple proofs. You can find an element $a\in {1,\ldots,p-1}$ such that $a^{p-1}=1 \mod p$ but $a^{(p-1)/2}\not=1\mod p$. See https://math.stackexchange.com/questions/837562/why-is-the-multiplicative-group-of-a-finite-field-cyclic – Jérémy Blanc Jun 29 '20 at 11:59