Quadratic Residue

$$x^2\equiv (x-n{)}^2(\mathrm{mod}n),0\le x\le n$$

In practice, it suffices to restrict the range to $0<x\le \lfloor \frac{n}{2}\rfloor$ because of the symmetry in $(x-n{)}^2\equiv x^2(\mathrm{mod}n)$:

$$1\mapsto 9(\mathrm{mod}10)$$ $$2\mapsto 8(\mathrm{mod}10)$$ $$3\mapsto 7(\mathrm{mod}10)$$ $$4\mapsto 6(\mathrm{mod}10)$$ $$5\mapsto 5(\mathrm{mod}10)$$

So for example, $x:=3$ in $(\mathrm{mod}10)$:

$$3^2\equiv (-7{)}^2(\mathrm{mod}10)$$ $$9\equiv 49(\mathrm{mod}10)$$

Read more:

[https://math.stackexchange.com/questions/1195366/all-the-solvable-congruences-x2-equiv-a-pmod-n-have-the-same-number-of-solu]

[https://en.wikipedia.org/wiki/Quadratic_residue]

[https://mathworld.wolfram.com/QuadraticResidue.html]