The sum of the quadratic residues modulo p

Step 1: there are exactly $m$ residues, represented by $1^2,\dots,m^2$. Any $x\not\equiv0$ is congruent to $\pm j$ for some $j\in\{1,\dots,m\}$, so $x^2\equiv j^2$. For $1\le i<j\le m$, $j^2-i^2=(j-i)(j+i)$ with $0<j-i<p$ and $0<j+i\le p-1<p$, so $p\nmid j^2-i^2$: the squares are distinct modulo $p$.

Step 2: the sum. Each residue $r$ is congruent to some $j^2$, so
$$\sum r\equiv\sum_{j=1}^mj^2=\frac{m(m+1)(2m+1)}6=\frac{p\,(p-1)(p+1)}{24}\pmod p,$$
using $m=\frac{p-1}2$, $m+1=\frac{p+1}2$, $2m+1=p$.

Step 3: divisibility. For a prime $p\ge5$, $p$ is odd and not a multiple of $3$, so $(p-1)(p+1)$ is divisible by $8$ and by $3$. Hence $24\mid(p-1)(p+1)$ and $\frac{p(p-1)(p+1)}{24}=p\cdot\frac{p^2-1}{24}$ is a multiple of $p$.

Check. $p=7$: residues $1,2,4$, sum $7$ ✓. $p=11$: residues $1,3,4,5,9$, sum $22$ ✓.

Lesson. To sum a set of residues, replace each by a convenient representative, here the squares $j^2$ with $j\le\frac{p-1}2$.