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Is still relevant or interesting be capable to bring a criteria in order to classifly quadratic extensions of $\mathbb{Q}$ based on the existence or not existence of non-trivial solutions of Fermat's cubic equation in the ring of integers belonging to those extensions? Considering d as the only number necessary to determine if $\mathbb{Q}[\sqrt{d}]$ has or not non-trivial solutions.

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  • $\begingroup$ See section 2 of this paper. $\endgroup$ – Wojowu Feb 26 at 20:38
  • $\begingroup$ I've already checked that paper. The result described, in general case, is not unknown for me. However I was not able to describe my idea clearly: i) Non-trivial solutions are searched in $Z[\omega _d]$, the ring of integers of $Q(\sqrt{d})$ $\endgroup$ – E. Pech Feb 26 at 23:41
  • $\begingroup$ ii) Criteria should be as simple as a modular congruence over Z. For example, "if $d\equiv 3(4)$ then there exist non-trivial solutions" Maybe I just can not see it, but I don't think Section 2 accomplish any of these points. $\endgroup$ – E. Pech Feb 26 at 23:49
  • $\begingroup$ Than you for all your help. $\endgroup$ – E. Pech Feb 26 at 23:50
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I don't want to toot my own horn, but I coauthored a paper on this topic with Marvin Jones. One direction of our result is conditional on the Birch and Swinnerton-Dyer conjecture (see the first remark on page 3). Basically, the Fermat cubic $x^3 + y^3 = z^3$ is isomorphic to $E : y^2 + y = x^3 - 7$. Since this curve has rank zero over $\mathbb{Q}$, the existence or non-existence of solutions in $\mathbb{Q}(\sqrt{d})$ is equivalent to the quadratic twist $E_{d}$ having positive rank or not over $\mathbb{Q}$. One can then give a criterion for this by relating $L(E_{d},1)$ to the Fourier coefficients of weight $3/2$ modular forms. The method is very similar to Tunnell's solution of the congruent number problem.

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    $\begingroup$ (though -- as with Tunnell's criterion -- in one direction part of the result is still conditional on the conjectural BSD criterion for positive rank.) $\endgroup$ – Noam D. Elkies Feb 27 at 1:22

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