 Quadratic nonresidue

A nonzero congruence class modulo an odd prime is a quadratic nonresidue when it is not a square. In the convention of quadratic residue, zero belongs to neither class. Thus $3,5,6$ are the quadratic nonresidues modulo seven, while $2,6,7,8,10$ are the quadratic nonresidues modulo eleven. No integer $x$ satisfies $x^2\equiv 3(\mathrm{mod}7)$, as the complete square map under quadratic residue shows.

A quadratic nonresidue is not an arithmetic dead end: multiplying by any fixed quadratic nonresidue $n$ exchanges the quadratic residues and the quadratic nonresidues. If $nq$ were a square for a square $q$, division by $q$ would make $n$ a square. Hence $n{Q}_p$ consists entirely of quadratic nonresidues; it has $(p-1)/2$ elements, because multiplication by nonzero $n$ is a bijection. It therefore equals the entire other coset of the quadratic residue subgroup. Modulo seven, $3\{1,2,4\}=\{3,6,5\}$.

The product of two quadratic nonresidues is consequently a quadratic residue: write them as $n{q}_1,n{q}_2$ in that same coset, and their product is $n^2{q}_1{q}_2$, a square. For example, $3\cdot 5\equiv 1(\mathrm{mod}7)$, whereas $2\cdot 3\equiv 6(\mathrm{mod}7)$ remains a quadratic nonresidue. The multiplicativity of the Legendre symbol packages these rules as ordinary sign multiplication.