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

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 $$\operatorname{rank}(E_D/K)=2$$ ?'

• Sorry, something seems off: when you say infinitely many twists $E_D$, the parameter you can vary is $D$, right? But in your post, $D$ seems to be fixed... Commented Jul 2, 2023 at 11:39
• Sorry, $D$ varies under the condition $D≡5mod8$. Commented Jul 2, 2023 at 11:42
• There is a standard strategy to produce points on twists: just plug in some value for $x$, and write $x^3+17x$ as $Dy^2$ with $D \in \mathbf Z$ squarefree and $y \in \mathbf Q$. So the question is whether you can choose $x$ in such a way that $D \equiv 5 \pmod 8$ and such that $(x,y)$ is not a torsion point. One way to force points not to be torsion is by Nagell–Lutz, which requires plugging in non-integer values for $x$ with this method. Commented Jul 2, 2023 at 11:46
• Oops, the equation $Dy^2 = x^3+Ax+B$ is a formula for the quadratic twist if $AB \neq 0$, but for the $j = 1728$ curve (i.e. $B = 0$) there are more twists (see Silverman, Ch. X, Prop. 5.4). Does "quadratic twist" for you mean the generic formula $y^2 = x^3+17D^2x$, or the special formula $y^2 = x^3+17Dx$ for the $j = 1728$ case? Commented Jul 2, 2023 at 13:20
• Thank you very much for comments. Yes, $E_d : y^2=x^3+17D^2x$. What I should do is to prove ''for all $D≡5mod8$, there are infinitely many $(x,y)$ such that $(x,y)\in E_d-(E_d)_{tor}$''. Torsion points should have integral coordinates, so it is enough to find no-integral $(x,y)$ which satisfies $y^2=x^3+D^2x$. But are there infinitely many non integral points in $y^2=x^3+D^2x$ for all $D≡5mod8$ ? Commented Jul 2, 2023 at 13:30

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 $$D \equiv 5 \pmod 8$$. $$\square$$

• How you take $Y$ as rational number ? For $X=m/4n$, I couldn't find the reason why $(X,Y)$ is a rational point of $E$. Commented Jul 3, 2023 at 10:15
• Any positive rational number $z$ (in this case $z = X^3+17X$) can be uniquely written as $z = DY^2$ where $D \in \mathbf Z$ is squarefree and $Y \in \mathbf Q$. Indeed, $D$ has to be the product of all primes $p$ with $v_p(z) \equiv 1 \pmod 2$, and then $z/D$ is a square in $\mathbf Q$. (If we did not care that $D$ is squarefree and integral, we would just take $Y=1$.) I'm just saying that every class in $\mathbf Q^\times/\mathbf Q^{\times 2}$ has a unique squarefree integer representative. Commented Jul 3, 2023 at 12:44
• Also recall that the equation $DY^2 = X^3+17X$ is equivalent to $y^2 = x^3+17D^2x$. Commented Jul 3, 2023 at 12:49
• The point is that I choose $x$ first and then choose $y$ and $D$ simultaneously. This is implicitly also what's going on in the other answer. But the other answer doesn't explain why the different values of $(x,y)$ cannot accidentally all give the same value of $D$. Commented Jul 3, 2023 at 15:47

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.

• Thank you very much. I try to prove without parity conjecture with the help of paper you provided. Commented Jul 3, 2023 at 5:57
• Could you erabolate why parity conjecture implies $'rank(E_D/\Bbb{Q})\ge 1 \implies rank(E_D/\Bbb{Q})\ge 2$ '? Commented Jul 13, 2023 at 2:28