Number of Quadratic Residues of Prime

Theorem
Let $$p$$ be an odd prime.

Then $$p$$ has $$\frac {p-1} 2$$ quadratic residues and $$\frac {p-1} 2$$ quadratic non-residues.

The quadratic residues are congruent modulo $p$ to the integers $$1^2, 2^2, \ldots, \left({\frac {p-1} 2}\right)$$.

Proof
The quadratic residues of $$p$$ are the integers which result from the evaluation of the squares $$1^2, 2^2, \ldots, \left({p-1}\right)^2$$ modulo $$p$$.

But $$r^2 = \left({-r}\right)^2$$ and so these $$p - 1$$ integers fall into congruent pairs modulo $$p$$, namely:

$$ $$ $$ $$

Therefore each quadratic residue of $$p$$ is congruent modulo $p$ to one of the $$\frac {p-1} 2$$ integers $$1^2, 2^2, \ldots, \left({\frac {p-1} 2}\right)$$.

Note that as $$r^2 \not \equiv 0 \pmod p$$ for $$1 \le r < p$$, the integer $$0$$ is not among these.

All we need to do now is show that no two of these integers are congruent modulo $p$.

So, suppose that $$r^2 \equiv s^2 \pmod p$$ for some $$1 \le r \le s \le \frac {p-1} 2$$.

What we are going to do is prove that $$r = s$$.

Now $$r^2 \equiv s^2 \pmod p$$ means that $$p$$ is a divisor of $$r^2 - s^2 = \left({r + s}\right) \left({r - s}\right)$$.

From Euclid's Lemma we see that either $$p \backslash \left({r + s}\right)$$ or $$p \backslash \left({r - s}\right)$$.

$$p \backslash \left({r + s}\right)$$ is impossible as $$2 \le r - s \le p - 1$$.

As for $$p \backslash \left({r - s}\right)$$, as $$0 \le r - s < \frac {p-1} 2$$, that can happen only when $$r - s = 0$$ or $$r = s$$.

So there must be exactly $$\frac {p-1} 2$$ quadratic residues, and that means there must also be exactly $$\frac {p-1} 2$$ quadratic non-residues.