Infinitely many elliptic curve with twist rank more than $1$ in specific case

Question

Let $E/\Bbb{Q}$ be an elliptic curve. Let $D$ be a square free negative integer.

It is conjectured that 50％ of twist of elliptic curve $E_D$ has rank $0$ and $50％$ has rank $1$.

But is some particular case known or conjectured?

I'm particularly interested in the case $E:y^2=x^3+17x$. Are there infinitely many twists $E_D$( $D≡5\bmod8$) which has $\operatorname{rank}(E_D/K)\ge 1$ ?

Reference is also appreciated. Thank you for your help.

P.S.

Thanks to Nulhomologous, by parity conjecture, what I should ask is 'Are there infinitely many twists $E_D(D≡5\bmod8$) which has $\operatorname{rank}(E_D/K)=2$ ?'

Answer 1

Let me turn my comment into an answer. There are indeed infinitely many such twists with a non-torsion point. By Nagell–Lutz, it suffices to produce infinitely many different squarefree integers $D \equiv 5 \pmod 8$ for which there exist $x,y \in \mathbf Q$ such that $y^2 = x^3+17D^2x$ and $x$ and $y$ are not both integers. Writing $X = Dx$ and $Y = D^2y$, we get the equivalent equation $DY^2 = X^3+17X$ (but beware that the Nagell–Lutz criterion does not apply in these coordinates).

One way to produce values of $D$ is as follows: there are infinitely many primes $p > 5$ such that $-17$ is a square modulo $p$. For such a prime $p$, pick $m \in \mathbf N$ such that $v_p(m^2+17) = 1$ and $m \equiv 5 \pmod 8$. Pick $n \in \mathbf N$ with $4n \equiv 1 \pmod p$ and $n \equiv 1 \pmod 8$. Set $X = \tfrac{m}{4n}$, and write $X^3+17X$ as $DY^2$ with $D \in \mathbf Z$ squarefree and $Y \in \mathbf Q$. Finally, set $x = \tfrac{X}{D}$ and $y = \tfrac{Y}{D^2}$.

We claim that $(x,y)$ is a non-torsion point on $E_D$, that $p \mid D$, and that $D \equiv 5 \pmod 8$. Since we can carry out this process for infinitely many primes $p$, we conclude that the numbers $D$ obtained this way also form an infinite set. It is clear that $x$ is not an integer, so $(x,y)$ cannot be a torsion point.

To see that $p \mid D$, note that $v_p(DY^2) = v_p(X^3+17X) = v_p(X^2+17) = 1$, so $v_p(D) = 1$ by definition of $D$. To see that $D \equiv 5 \pmod 8$, note that $$DY^2 = X^3+17X = X(X^2+17) = \tfrac{m}{4n}\cdot \tfrac{m^2+272n^2}{16n^2}.$$ Since $m$ and $n$ are odd, we see that $v_2(Y^2) = -6$ and $v_2(D) = 0$. Factoring out all even denominators, the above equation reads $$D \cdot (2^3Y)^2 = \tfrac{m}{n} \cdot \tfrac{m^2+272n^2}{n^2}.$$ Since $2^3Y$ is a unit in $\mathbf Z_{(2)}$, its square is 1 modulo 8. Since $m$ and $n$ are odd, we get $\tfrac{m^2+272n^2}{n^2} \equiv 1 \pmod 8$, and we have $\tfrac{m}{n} \equiv 5 \pmod 8$ by the choice of $m$ and $n$. Putting everything together gives $D \equiv 5 \pmod 8$. $\square$

Answer 2

I answer the original question, about the existence of infinitely many $D$'s, $D\equiv 5 \pmod{8}$, such that the quadratic $D$-twist of $E: y^2=x^3+17x$ has infinitely many points. As indicated, this should imply under the parity conjecture, that there are infinitely many such twists with rank $\ge 2$ (and even).

First of all, about the torsion of the twists of the curve $E$: it is always the group with two elements (over the rationals). The torsion point is $(0,0)$. This can be deduced by showing that the twists cannot contain a point of order 4 or a point of order 6.

So we only need to show there are infinitely many $D$'s with a rational point with $x\ne 0$ in $E_D$ and $D\equiv 5 \pmod{8}$.

Note that one can do a quadratic twist with any non-zero rational number $z$, but there is a unique $D$ square free integer such that $z/D$ is a square; then the quadratic twist with respect to $z$ and with respect to $D$ are isomorphic.

Consider the polynomial $f=y^2-x^3-17z^2x$ where the variables are $x$, $y$ and $z$, over the field of rational numbers. This defines an affine surface which is rational and easily parametrizable (the points with $x=0=y$ are special, but we are not interested with them).

A possible parametrization is $x=t^2/(s^2 + 17)$, $y=t^3/(s^3 + 17s)$ and $z=t^2/(s^3 + 17s)$. Now your question is equivalent to the question: there exists infinitely many integers $D$, congruent to $5$ modulo 8, such that $D= u^2t^2/(s^3 + 17s)$ for some $u$, $t$ and $s\in \mathbb{Q}$? It is clear that this is equivalent to just ask if $D$ can be written as $D= u^2(s^3 + 17s)$ for some $u$, $s\in \mathbb{Q}$?

Or, more directly and elementarily, one can see that the point $(x,y)=(1/(s^2 + 17),1/(s^3 + 17s))$ is a solution of the equation $y^2=x^3+17/(s^3 + 17s)^2x$.

Writting $s=\frac ab$, with $a$ and $b$ integers, then $s^3 + 17s$ is equal to $b(a^3+17ab^2)$ modulo squares. Notice that if $b(a^3+17ab^2)$ is congruent to $5$, and $D$ is a square free integer equal to $b(a^3+17ab^2)$ modulo squares, then $D\equiv 5 \pmod{8}$. But this never happens, as $b(a^3+17ab^2)$ is always even.

Hence, we can look for $b(a^3+17ab^2) \equiv 20 \pmod{32}$, which will imply that, if $D$ is a square free integer equal to $b(a^3+17ab^2)$ modulo squares, then $D\equiv 5 \pmod{8}$.

But it is easy to find infinitely many solutions of this equation: for example, if $b\equiv 4\pmod{32}$ and $a\equiv 5 \pmod{32}$.

p.s. It could be possible to prove that there are infinitely many quadratic twists of this curve that have at least two independent points, and may be deduce the same for the quadratic twists with the condition you are interested (without using any conjecture), using the ideas given by Mestre and by Steward and Top. See the paper "On Ranks of Twists of Elliptic Curves and Power-Free Values of Binary Forms" and the references therein.