For any integer $n$, $n^2$ is congruent to $0$ or $1$ modulo $3$

Question

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?

Answer 1

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 $[2]_3$.

Answer 2

Fermat's little theorem says that $a^2\cong 1\pmod3$, if $a\not\cong0\pmod 3$.

But this can be done by hand. We can check that $0^2\cong0\pmod 3$, $1^2\cong1\pmod 3$, and $2^2\cong4\cong1 \pmod3$. There are only three cases to check.