For any integer $n$, $n^2$ is congruent to $0$ or $1$ modulo $3$ I have tried out to prove it by its contrapositive.
Therefore now I am proving that  there is no integer whose square is congruent to $2 \pmod 3$.
Consider the case $n^2 \equiv 2 \pmod 3$
$ n^2 = 3k +2$
I am stuck to show $\sqrt{3k+2}$ is not a integer. Can anyone provide some hint to me?

Observe: $[0]_3^2=[0\times 0]_3=[0]_3$ and $[\pm 1]^2_3=[(\pm 1)\times (\pm 1)]_3=[1]_3$; since $\{-1,0,1\}$ is a complete system of residues modulo three, all equivalence classes $[a]_3$ modulo three are accounted for, and none of which square to $[2]_3$.
