If $d$ is divisible by a prime $p \equiv 3 \pmod{4}$. show that the equation $x^2-dy^2=-1$ has no solution.

So far I have learn only positive Pell's equation but not negative Pell's equation. We know that in positive Pell's equation, it always has a solution but not the case for negative Pell's equation.

Can anyone give some hints on how to tackle this question?

  • $\begingroup$ Do you know what quadratic residues are, Legendre symbol? $\endgroup$
    – Will Jagy
    Dec 1, 2013 at 4:35
  • $\begingroup$ @WillJagy: Yup, I do learn that. So the negative pell's equation have no solution iff the Legendre symbol $(\frac{-1}{d})=-1$? $\endgroup$
    – Idonknow
    Dec 1, 2013 at 4:37
  • $\begingroup$ When $p \equiv 3 \pmod 4$ and $x^2 \equiv -1 \pmod p$, what is $x^{p-1}\pmod p$ ? $\endgroup$ Dec 1, 2013 at 4:53
  • $\begingroup$ @achillehui: $x^{p-1} \equiv -1 (\mod p)$? $\endgroup$
    – Idonknow
    Dec 1, 2013 at 4:56
  • $\begingroup$ Yup, and this contradicts with Fermat's little theorem. $\endgroup$ Dec 1, 2013 at 5:00

1 Answer 1


Suppose a solution exists. Then reducing mod $p$ we find that $x^2 \equiv -1 \bmod p$. Then $\left(\frac{-1}{p}\right) = 1$. However $p\equiv 3 \bmod 4$ so this is a contradiction.


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