# $(x \equiv k^2 \mod 3) \iff x \equiv 1 \mod 3$

Is it true that if 3 does not divide $x$,

$$x\equiv k^2\mod 3 \iff x\equiv 1 \mod 3$$

If the above statement is correct , There are two parts to prove

$$x\equiv k^2\mod 3 \implies x\not\equiv 0 \mod 3$$ $$x\equiv k^2\mod 3 \implies x\not\equiv 2 \mod 3$$

How to prove them ?

• Your question is not clear, because $3\not|x \implies x\not\equiv0\pmod3$, i.e., if $3$ does not divide $x$, then obviously $x\not\equiv0\pmod3$. – barak manos Sep 10 '14 at 5:55
• @barakmanos yeah , thanku . – hanugm Sep 10 '14 at 6:13

You can see from this analysis that the only possible values for $k^2$ are $0,1\pmod{3}$, which is what you wanted to show.