Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a continuation of this previous problem I asked about. From Ireland and Rosen's Number theory book(ch.11 ex#12)

EDITED: Let $C_{1}$ be the curve $y^2=x^{3}-Dx$ over $\mathbb{F}_{p}$, where $D\in \mathbb{F}_{p}$ and $D \not= 0$

If $p=3\mod{4}$, show that the number of projective points on $C_{1}$ is just $p+1$. If $p=1\mod{4}$, show that the answer is $p+1+\overline{\chi (D)} J(\chi,\chi^{2})+\chi (D)\overline{J(\chi,\chi^{2})}$ where $\chi$ is a character of order 4 on $\mathbb{F}_{p}$

share|cite|improve this question
up vote 3 down vote accepted

I'll do the easy part. Note that $x^3-Dx$ is an odd function of $x$. If $p\equiv3\pmod4$, then $r$ is a square modulo $p$ if and only if $-r$ isn't. So for half the nonzero values of $x$, $y^2=x^3-Dx$ has two solutions, and for the other half (the negatives of the first half), no solutions. Then there's exactly one solution when $x=0$, and then there's the point at infinity, so you get $p+1$.

EDIT: I won't work through all the detail of the hard part, but maybe this will steer you in the right direction.

The previous exercise tells you the number of points on your curve if you know the number on $u^2-v^4=4D$. Now that's not a million miles from what's toward the bottom of page 167, where the number of points on $w^2+z^4=1$ is found, using Theorem 5 of Chapter 8, to be $$p+J(\rho,\chi)+J(\rho,\chi^2)+J(\rho,\chi^3)$$ where $\rho$ is the quadratic, $\chi$ the quartic, character. Then they show how to simplify that to $$p-1-\pi-\overline{\pi}$$ where $\pi=-J(\rho,\chi)$. That looks a lot like what you want, so I think the key is seeing what Theorem 5 of Chpter 8 says about the equation and then following the simplifying steps on page 167.

share|cite|improve this answer
Thank you Gerry for this great answer. – Edison Apr 21 '12 at 3:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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