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 ?

  • 1
    $\begingroup$ Your question is not clear, because $3\not|x \implies x\not\equiv0\pmod3$, i.e., if $3$ does not divide $x$, then obviously $x\not\equiv0\pmod3$. $\endgroup$ – barak manos Sep 10 '14 at 5:55
  • $\begingroup$ @barakmanos yeah , thanku . $\endgroup$ – hanugm Sep 10 '14 at 6:13

A cleaner way of stating this would be, that the only squares modulo 3 are 0 and 1. We can prove this, by squaring each of the three residues modulo 3:

\begin{align*} 0^2 &\equiv 0\pmod{3} \\ 1^2 &\equiv 1\pmod{3} \\ 2^2 \equiv 4 &\equiv 1\pmod{3} \end{align*}

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.

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