0
$\begingroup$

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 ?

$\endgroup$
  • 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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.