Exercise 3.1.14 (The sum of the quadratic residues is divisible by $p$)

Proof. Every quadratic residue modulo $p$ is by definition congruent to $1^2,2^2,\dots ,{(p-1)}^2$. Moreover, $a^2\equiv {(p-a)}^2(\mathit{mod}p)$ for every $a$, thus every quadratic residue modulo $p$ is by definition congruent to $1^2,2^2,\dots ,{((p-1)∕2)}^2$.

Note that the classes of the $(p-1)∕2$ integers are distinct, otherwise there are less than $(p-1)∕2$ classes of quadratic residues. But Problem 12 shows that there are exactly $(p-1)∕2$ such classes. So every quadratic residue modulo $p$ is congruent to one and only one integer in the set $\{1^2,2^2,3^2,\dots ,{((p-1)∕2)}^2\}$.

This shows that the sum $S$ of residues modulo $p$ satisfies

Using

we obtain, for $n=(p-1)∕2$,

where $\frac{(p^2-1)p}{24}={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^{(p-1)∕2}{j}^2$ is an integer. Thus $24\mid (p^2-1)p$, and $24\wedge p=1$, since $p>3$ by hypothesis, thus $24\mid p^2-1$ is true for every prime $p>3$, so

If $p>3$, the sum of the quadratic residues is divisible by $p$. □