Algebra with congruence classes.

Question

Consider $x^2=[2]_7$ where $[2]_7$ is the congruence class of $a$ modulo $m$, or $$[2]_7=\{x\in\mathbb{Z} \ | x\equiv 2 \ (\text{mod} \ 7)\}.$$ It is immediately obvious, by inspection, that $x=[3]_7$ and $x=[4]_7$, but is there any way to analytically attain these solutions?

Answer 1

Just compute the squares in $\mathbf Z/7\mathbf Z=\bigl\{0,\pm 1,\pm 2,\pm3\bigr\}$: $$0^2=0,\quad (\pm1)^2=1,\quad (\pm2)^2=-3,\quad (\pm3)^2=2,$$ hence the solutions are $3$ and $-3\equiv 4\mod7$.

Answer 2

$ x^2 \equiv 2 \iff 4x^2 \equiv 8 \iff (2x)^2 \equiv 1 $

$\iff$ $7$ divides $(2x)^2 - 1 = (2x-1)(2x+1)$ $\iff$ $7$ divides $2x-1$ or $7$ divides $2x+1$

$ \iff 2x \equiv \pm 1 \iff 8x \equiv \pm 4 \iff x \equiv \pm 4 $

In this case we are lucky that the inverse of $2$ is clearly a square.

In other cases, we might not be so lucky: if we replace $7$ by $23$, then the inverse of $2$ is $12$, not clearly the square of $14$.