16
$\begingroup$

Let $E: y^2 = x^3 + ax + b$ be an elliptic curve over $\mathbb{Q}$ with $a,b \in \mathbb{Z}$. Recall that any rational point $P = (x,y)$ can be written uniquely as $P = (u/d^2, v/d^3)$ with $u,v,d \in \mathbb{Z}, d > 0, \gcd(u,d) = \gcd(v,d) = 1$. We write $d(P)$ for the number appearing in the denominator.

The sequence $d(nP)$ for $n \in \mathbb{N}$ is the elliptic divisibility sequence associated to $P$. My question is about existence of quadratic non-residues in these sequences modulo all but finitely many primes $p$.

Let $P$ be a point of infinite order. Does there exists a finite set of primes $S$ such that for all $p \notin S$, there is $n \in \mathbb{N}$ such that $$\left( \frac{d(nP)}{p}\right) = -1 \quad ? \quad (*)$$

Remarks:

  1. It is known that the sequence $d(nP)$ contains only finitely many squares, by Theorem 1.1 of

    Everest, Reynolds, Stevens - On the denominators of rational points on elliptic curves.

    In particular, for all but finitely many $n$, one finds that $d(nP)$ is a quadratic non-residue modulo half of all primes. I want to know that as one varies $n$ and takes all these primes together, one only excludes finitely many primes.

  2. There is a slightly different definition of elliptic divisibility sequences in terms of division polynomials and recurrence sequences, which can differ from the above sequence by a sign and some small primes (this is the original definition of Ward). I'm also interested in the analogous problem for such sequences.

$\endgroup$
  • 1
    $\begingroup$ Is this known for Lucas sequences? $\endgroup$ – Dror Speiser Jul 27 at 20:15
  • $\begingroup$ Good question. I don't know. $\endgroup$ – Daniel Loughran Jul 28 at 8:22
  • $\begingroup$ Is it even known that the primes excluded have density zero? $\endgroup$ – Stanley Yao Xiao Jul 30 at 12:03
  • 1
    $\begingroup$ @Stanley Yao Xiao up to some work reconciling the definition difference, I think so: combine the main result in Reductions of Points on Elliptic Curves by Akbary et al, with the main result in Character Sums with Division Polynomials by Shparlinski and Stange. $\endgroup$ – Dror Speiser Jul 30 at 17:08
  • 1
    $\begingroup$ It appears that for a fixed $E$ and $P$, the primes for which there is no $n$ so that $\left(\frac{d(nP)}{p}\right) = -1$ can be reasonably large. I just looked at the example of $E : y^{2} = x^{3} - 2x$ and $P = (2,2)$, and found that there is no such $n$ for $p = 17$, $257$, $1009$, $1361$, $26881$, and $141041$. $\endgroup$ – Jeremy Rouse Aug 5 at 2:06

Your Answer

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

Browse other questions tagged or ask your own question.