Under the abc conjecture, $X \ll_\epsilon D^{1+\epsilon}$. ("Infinite order" is irrelevant because for all but finitely many $D$ only 2-torsion points can have infinite order.) This also means that you can't use "Pell equation" tricks to make $X > D^{1+C}$ for any $C>0$. But (unless you believe that abc has been proved) it's probably very hard to prove $X \ll_\epsilon D^{1+\epsilon}$ unconditionally.

For the example of $y^2 = x^3 - x$, we can apply the abc conjecture to $(X^2-1) + 1 = X^2$. This has height $H = X^2$. If $X^3 - X = D Y^2$ then the conductor $N$ is at most $DY$, so $D = (DY)^2 / (DY^2) \leq N^2/(X^3-X) \sim N^2 / X^3,$ which is at least $(X^4)^{1-\epsilon} / X^3 = X^{1-4\epsilon}$ under ABC.

For other fixed $A,B$ such that $x^3 + aX + B$ has no repeated roots, we handle $DY^2 = X^3 + AX + B$ as follows. By Belyi's theorem there are relatively prime polynomials $a(x), b(x), c(x) \in {\bf Z}[x]$ such that:

• $a+b=c$,
• $a$ is a multiple of $X^3+AX+B$,
• $\deg(b)=\deg(c)=d$ for some positive integer $d$, and
• $\deg(a) < d$.

(We recover $a,b,c$ from a Belyi function $a/c$ sending $\infty$ and the roots of $x^3+Ax+B = 0$ to zero.) Now $a(X)+b(X)=c(X)$ with a bounded gcd; dividing by this gcd we get an abc triple of height $H \gg X^d$. Thus the abc conjecture implies $N \gg_\epsilon X^{d-\epsilon}$. But the repeated roots of $a,b,c$ make $N \ll X^{d+1}$ for all $X$; if moreover $X^3+AX+B = D Y^2$ then $N \leq X^{d+1}/Y$. Thus abc implies $Y \ll X^{1+\epsilon}$; again $DY^2 \sim X^3$ so it follows that $X$ cannot increase faster than $D^{1+o(1)}$.

Naturally this is yet another variation of the use of Belyi functions that I introduced in "ABC implies Mordell" back in 1991. I would not be surprised to learn that this application has already been noted before.

Under the abc conjecture, $X \ll_\epsilon D^{1+\epsilon}$. ("Infinite order" is irrelevant because for all but finitely many $D$ only 2-torsion points can have finite order.) This also means that you can't use "Pell equation" tricks to make $X > D^{1+C}$ for any $C>0$. But (unless you believe that abc has been proved) it's probably very hard to prove $X \ll_\epsilon D^{1+\epsilon}$ unconditionally.

For the example of $y^2 = x^3 - x$, we can apply the abc conjecture to $(X^2-1) + 1 = X^2$. This has height $H = X^2$. If $X^3 - X = D Y^2$ then the conductor $N$ is at most $DY$, so $D = (DY)^2 / (DY^2) \leq N^2/(X^3-X) \sim N^2 / X^3,$ which is at least $(X^4)^{1-\epsilon} / X^3 = X^{1-4\epsilon}$ under ABC.

For other fixed $A,B$ such that $x^3 + aX + B$ has no repeated roots, we handle $DY^2 = X^3 + AX + B$ as follows. By Belyi's theorem there are relatively prime polynomials $a(x), b(x), c(x) \in {\bf Z}[x]$ such that:

• $a+b=c$,
• $a$ is a multiple of $X^3+AX+B$,
• $\deg(b)=\deg(c)=d$ for some positive integer $d$, and
• $\deg(a) < d$.

(We recover $a,b,c$ from a Belyi function $a/c$ sending $\infty$ and the roots of $x^3+Ax+B = 0$ to zero.) Now $a(X)+b(X)=c(X)$ with a bounded gcd; dividing by this gcd we get an abc triple of height $H \gg X^d$. Thus the abc conjecture implies $N \gg_\epsilon X^{d-\epsilon}$. But the repeated roots of $a,b,c$ make $N \ll X^{d+1}$ for all $X$; if moreover $X^3+AX+B = D Y^2$ then $N \ll X^{d+1}/Y$. Thus abc implies $Y \ll X^{1+\epsilon}$; again $DY^2 \sim X^3$ so it follows that $X$ cannot increase faster than $D^{1+o(1)}$.

Naturally this is yet another variation of the use of Belyi functions that I introduced in "ABC implies Mordell" back in 1991. I would not be surprised to learn that this application has already been noted before.