Theorem. Let $p$ be a positive odd prime number. Then the integers

(1)

Proof. Firstly, the numbers (1), being squares, are quadratic residues modulo $p$ . Secondly, they are incongruent, because a congruence $a^2 \equiv b^2 \pmod{p}$ would imply $$p \mid a\!+\!b \quad \mbox{or} \quad p \mid a\!-\!b,$$ which is impossible when $a$ and $b$ are different integers among $1,\,2,\,\ldots,\,\frac{p\!-\!1}{2}$ . Third, if $c$ is any quadratic residue modulo $p$ , and therefore the congruence $x^2 \equiv c \pmod{p}$ has a solution $x$ , then $x$ is congruent with one of the numbers $$\pm1,\,\pm2,\,\ldots,\,\pm\frac{p\!-\!1}{2}$$ which form a reduced residue system modulo $p$ (see absolutely least remainders). Then $x^2$ and $c$ are congruent with one of the numbers (1).