Exercise 14.4.5

Proof. (a) Consider the map

For all $s,t\in {𝔽}_p^{\text{∗}}$, $\phi (\mathit{st})={(\mathit{st})}^2=s^2{t}^2=\phi (s)\phi (t)$, thus $\phi$ is a group homomorphism.

Not that, $\mathrm{Im}(\phi )=\{s^2\mid s\in {𝔽}_p^{\text{∗}}\}=S\setminus \{0\}$. Moreover $s\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ iff $s^2=1$, that is $(s-1)(s+1)=0$. Since ${𝔽}_p$ is a field, this is equivalent to $s=1$ or $s=-1$, thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{-1,1\}$, where $-1\ne 1$ since $p>2$. By the first Isomorphism Theorem,

This shows that there are $(p-1)∕2$ squares in ${𝔽}_p^{\text{∗}}$. If we add the square $0=0^2$, we obtain

(b) The two maps $u,t_a:{𝔽}_p\to {𝔽}_p$ defined for all $s\in {𝔽}_p$ by $u(s)=-s$, and $t_a(s)=a+s$ are bijective (since $u\circ u=1_{{𝔽}_p}$ and $t_a\circ t_{-a}=t_{-a}\circ t_a=1_{{𝔽}_p})$. Therefore

is bijective.

Since $S^{\prime }=\{f(s)\mid s\in {𝔽}_p\}=f(S)$,

Reasoning by contradiction, if $S\cap S^{\prime }=\varnothing$, then the inclusion $S\cup S^{\prime }\subset {𝔽}_p$ shows that

and $p+1\le p$ gives a contradiction. Therefore $S\cap S^{\prime }\ne \varnothing$, so that we can find some $c\in {𝔽}_p$ such that $c\in S\cap S^{\prime }$. Such an element $c$ verifies $c=s^2$ and $c=a-t^2$ for some $s,t\in {𝔽}_p$. This proves that the equation $s^2+t^2=a$ has a solution $(s,t)\in {𝔽}_p\times {𝔽}_p$. □