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Consider the Legendre family of elliptic curves $$E_a: y^2=x(x-1)(x-a).$$ Let $p$ be an odd prime.

QUESTION. Is the following true? If $p\equiv 3\pmod4$ then number of solutions to $E_2$ over the finite field $\mathbf{F}_p$ is $$\#(E_2/\mathbf{F}_p)\equiv 0 \pmod p.$$

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    $\begingroup$ Yes, because substituting x+1 for x identifies E_2 with the curve y^2 = x^3 - x for which this result is well known. (Indeed the count is exactly p. Not p+1, because as stated the question excludes the point at infinity.) $\endgroup$
    – Noam D. Elkies
    Commented Mar 8 at 19:01
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    $\begingroup$ That curve has complex multiplication by $\mathbb{Z}[i]$ so, $a_p=0$ for all primes $p\equiv 3 \pmod{4}$. That means that the number of affine points is $p$ and the group order is $p+1$. $\endgroup$
    – Chris Wuthrich
    Commented Mar 8 at 19:03
  • $\begingroup$ Thanks to both Noam D. Elkies and Chris Wuthrich. $\endgroup$
    – T. Amdeberhan
    Commented Mar 20 at 14:48

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Corresponding sums of Legendre symbols are know as Jacobsthal sums $$J(u)=\sum_{x\mod p}\left(\frac {x^3+ux}p\right).$$ They are equal to $0$ for prime $p\equiv 3\pmod4$. If $p\equiv 1\pmod4 $, $\left(\frac ap\right)=1$ and $\left(\frac bp\right)=-1$, then $J(a)$, $J(b)$ are even and $$\left(\frac {J(a)}2\right)^2+\left(\frac {J(b)}2\right)^2=p.$$

The original Jacobsthal's article is available here.

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  • $\begingroup$ I think that this (or something easily equivalent) was already known to Gauss as part of hiw work on quartic (a.k.a. "biquadratic") reciprocity. $\endgroup$
    – Noam D. Elkies
    Commented Mar 11 at 21:14

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