2
$\begingroup$

For an odd prime $p$, prove that the quadratic residues of $p$ are congruent modulo $p$ to the integers $$1^2,2^2, 3^2,\ldots, \left(\dfrac{p-1}{2}\right)^2$$

I know Euler's criterion but not sure how to start the proof. Any help is appreciated. Thanks!

$\endgroup$
2
$\begingroup$

Quadratic residues modulo $p$ consists of the numbers $1^2,2^2,\ldots ,(p-1)^2$ by definition. But these $p-1$ squares are pairwise congruent because of $x^2\equiv (x-p)^2 \bmod p$. So it suffices to take $1^2,2^2,\ldots ,\left(\frac{p-1}{2}\right)^2$.

$\endgroup$
  • $\begingroup$ Ahh so the first $(p-1)/2$ squares are congruent to the last $(p-1)/2$ squares... nice xD got it ! ty =) $\endgroup$ – rrr Feb 19 '15 at 21:12

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.