Let $K$ be a number field and let $\mathcal O\_K$ be the ring of integers. Following [this paper](https://www.emis.de/journals/JTNB/2005-3/article02.pdf "Division-ample sets and the Diophantine problem for rings of integers") of Cornelissen, Pheidas, and Zahidi, a key ingredient needed to show that Hilbert's tenth problem has a negative solution over $\mathcal O\_K$ is an elliptic curve $E$ defined over $K$ with rank$(E(K))=1$.


Recently [Mazur and Rubin](https://arxiv.org/pdf/0904.3709v2 "Ranks of twists of elliptic curves and Hilbert's Tenth Problem") have shown that such a curve exists assuming the Shafarevich-Tate conjecture for elliptic curves over number fields. They actually use a weaker, but still inaccessible hypothesis (See conjecture $ШT\_2$).


If you wanted to eliminate the need for this hypothesis you would have to write a proof that simultaneously demonstrated that rank$(E(K))=1$ for infinitely many pairs $(K,E)$ where $E$ is an elliptic curve defined over $K.$ This raises (as opposed to begs) the easier question:



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> Can you show unconditionally that rank$(E(\Bbb Q)) = 1$ for infinitely many elliptic curves $E$ over $\Bbb Q$?
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It would appear that Byeon, Jeon, and Kim have done so in [this paper (probably need an institutional login)](https://doi.org/10.1515/CRELLE.2009.060 "Rank-one quadratic twists of an infinite family of elliptic curves"). Vatsal obtains a weaker result [here](https://personal.math.ubc.ca/%7Evatsal/research/der.PDF "Rank-one Twists of a Certain Elliptic Curve") that still does the job. Unfortunately both of these results invoke the fact that the BSD rank conjecture is true for elliptic curves over $\Bbb Q$ with analytic rank 1. Which won't help at present working over number fields.



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> Can anyone do the above **WITHOUT** invoking the proven part of the BSD rank conjecture or assuming any conjectures?
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Can you show rank E(Q) = 1 exactly for infinitely many elliptic curves E over Q without using BSD?
I don't think that this should be too hard: take a simple family of curves, such as 
$y^2 = x^3 + px$ or something similar, and choose $p$ from a certain set of residue classes to guarantee that the 2-Selmer group has rank 1. You can complete the proof either by invoking rather deep constructions using Heegner points, or finding a family for which conditions such as $p = a^4 + b^2$ (there are infinitely many primes of this form) give you a global point (choose your family in such a way that $b^2 = px^4 - y^4$ occurs as a 
principal homogeneous space in your standard 2-descent; see e.g. Silverman's book).
Sorry for being a little bit vague - a hard disk crash currently prevents me from looking at my own notes.


