Exercise 5.2

Proof. Let $N$ be the number of solutions of $x^2\equiv a(\mathit{mod}p)$.

$\text{∙}$ If $\left(\frac{a}{p}\right)=0$, then $p\mid a$, $a\equiv 0(\mathit{mod}p)$, so the unique solution of $x^2\equiv a=0$ is $x\equiv 0(\mathit{mod}p)$, so $N=1=1+\left(\frac{a}{p}\right)$.

$\text{∙}$ If $\left(\frac{a}{p}\right)=-1$, then $N=0=1+\left(\frac{a}{p}\right)$.

$\text{∙}$ If $\left(\frac{a}{p}\right)=1$, then $x^2\equiv a(\mathit{mod}p)$ has a solution $x_0$, and $x^2\equiv a(\mathit{mod}p)⟺x^2\equiv x_0^2(modp)\equiv p\mid (x-x_0)(x+x_0)\equiv x\equiv \pm x_0(modp)$, so $N=2=1+\left(\frac{a}{p}\right)$. □