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 $\operatorname{rank}(E_D/K)=2$ ?'
Best Answer
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 $D \equiv 5 \pmod 8$. $\square$