For every rational number $h$, does there exist a rational $t$ such that the rank of the elliptic curve $$Y^2=X^3 3h^2t^8X (h^4t^{16}+h^2t^8),$$ is positive?

$\begingroup$ I would appreciate any answers to this question. Is it right? It seems right, but how I can prove it? $\endgroup$ – mehdi baghalaghdam Jan 5 '17 at 16:24

2$\begingroup$ How does this problem arise in your work? $\endgroup$ – Daniel Loughran Jan 5 '17 at 16:59

$\begingroup$ We use this elliptic curve in elliptic curves cryptosystems.(Ecc) $\endgroup$ – mehdi baghalaghdam Jan 5 '17 at 17:52

1$\begingroup$ There might be a cheap trick for doing this  but on the other hand it could well be the case that giving an unconditional proof of this might be hard or even open. Sometimes you can do tricks involving showing that two distinct values of $t$ give two elliptic curves with different global signs in the functional equation and then you're done if you believe BSD, but perhaps you should make clear whether an answer which depends on standard conjectures is suitable from your point of view. You've asked lots of questions about these curves recently but the arithmetic of ell curves can be hard. $\endgroup$ – Kevin Buzzard Jan 5 '17 at 20:41

1$\begingroup$ Here are some people (including some experts) wondering whether something slightly more general is a reasonable conjecture: mathoverflow.net/a/63970/1384 . $\endgroup$ – Kevin Buzzard Jan 5 '17 at 23:11
I believe that one can expect an easy solution only if there are rational functions $X(h), Y(h), T(h)\in\mathbb Q(h)$ for a variable $h$ such that $Y(h)^2=X(h)^3 3h^2T(h)^8X(h) (h^4T(h)^{16}+h^2T(h)^8)$.
I don't know how to decide easily if there are such functions. Still, the curve has an interesting feature: For $h=1$ it has positive rank over the rational function field $\mathbb Q(t)$, since $(X,Y)=(t^2+t^4+t^6,t^3+t^5+t^7+t^9)$ is a point on the curve, which is not torsion by NagellLutz.

$\begingroup$ Dear Prof Peter Mueller. I appreciate your reply and your answer. thank you.. It is great. How did you find this answer? Please more explain about why this point is not torsion? $\endgroup$ – mehdi baghalaghdam Jan 7 '17 at 2:04

1$\begingroup$ @mehdi: I was playing a little with the computer algebra system Sage, which can compute generators of the MordellWeil group of elliptic curves over the rationals. The point over $\mathbb Q(t)$ was found by polynomial interpolation of the various points over $\mathbb Q$. If the given point were a torsion point, then it would be one upon specializing $t$, for instance setting $t=2$. So $p=(84, 680)$ were a torsion point on $Y^2 = X^3  768X  65792$, and so would be $3p$. However, $3p=(809236/729,727724440/19683)$ has nonintegral entries, contrary to NagellLutz. $\endgroup$ – Peter Mueller Jan 7 '17 at 16:20