Exercise 5.3

Proof. Here $p$ is an odd prime number, and $p\nmid a$. Let $N$ be the number of solutions of $a{x}^2+\mathit{bx}+c\equiv 0(\mathit{mod}p)$

For $\bar{x}\in {𝔽}_p=\mathbb{Z}∕\mathit{p\mathbb{Z}}$,

Let $\mathrm{\Delta }=b^2-4\mathit{ac}$. Then $N$ is the number of solutions of ${\left(\bar{x}+\frac{\bar{b}}{2\bar{a}}\right)}^2-\frac{\bar{\mathrm{\Delta }}}{4{\bar{a}}^2}=\bar{0}$ in ${𝔽}_p$. As in Ex.5.2, $N=1$ if $\bar{\mathrm{\Delta }}=\bar{0}$, $N=0$ if $\bar{\mathrm{\Delta }}$ is not a square in ${𝔽}_p^{\text{∗}}$, otherwise $\bar{\mathrm{\Delta }}={\delta }^2,\delta \in {𝔽}_p^{\text{∗}}$, and the solutions are $\bar{x}=(-\bar{b}\pm \bar{\delta })∕2\bar{a}$, so $N=2$. In the three cases, $N=1+\left(\frac{\mathrm{\Delta }}{p}\right)$. □