Timeline for Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2(\sqrt{-5})$?
Current License: CC BY-SA 4.0
11 events
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| Jul 14, 2023 at 19:26 | comment | added | Chris Wuthrich | As noted in this question, my previous comment is misleading due to an error in Simon's script. | |
| Jul 6, 2023 at 8:32 | comment | added | Chris Wuthrich |
For the original question of a $2$ descent over $\mathbb{Q}(\sqrt{-5})$. This is implemented. For instance sage: E = EllipticCurve([17,0]) sage: K.<t> = QuadraticField(-5) sage: EK = E.base_extend(K) sage: EK.simon_two_descent(verbose=4) will return details on how that script proves that Sha is non-trivial (though it doesn't determine it completely. The code is public so you gather there how one calculates these things efficiently. There is a lot of literature on these equations related to such implementations.
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| Jul 6, 2023 at 7:59 | answer | added | Achim Krause | timeline score: 4 | |
| Jul 6, 2023 at 7:00 | comment | added | Achim Krause | For any given $x$, checking whether $-1-17x^4$ is a square in $K$ amounts to checking whether it lies in $(K^{\times})^2$. I'm not claiming the whole solution is up to squares. | |
| Jul 6, 2023 at 5:57 | comment | added | Duality | @Achim Krause. The soluton is not up to ${K^{\times}}^2$ I think. I don't understand the reason why you can calculate up to ${K^{\times}}^2$. | |
| Jul 6, 2023 at 5:53 | comment | added | Duality | @KContad Thank you for your advice. Indeed, this calculation occurred to calculate Selmer group over quadratic fields. But to judge whether $x^4+y^2=-1$ has solution or not in $\Bbb{Q}(\sqrt{-5})$ seems difficult as the same as original one. | |
| Jul 6, 2023 at 5:52 | history | edited | Duality | CC BY-SA 4.0 |
added 130 characters in body
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| Jul 5, 2023 at 17:30 | comment | added | Achim Krause | Here's a nice general trick: Since solvability of the equation $ay^2 = b + cx^4$ is equivalent to finding $x$ such that $a^{-1}(b + cx^4)$ is a square, and in a $p$-adic number field $K^\times / (K^\times)^2$ is finite, for any given $K$ it suffices to know an element up to an explicitly computable order to decide whether it's a square. For $x$ of small valuation, this boils down to deciding whether $a^{-1}c$ is square, and for $x$ of not-too-small valuation this leaves finitely many values of $x$ to check. This can be done quite mechanically. | |
| Jul 5, 2023 at 17:19 | comment | added | KConrad | Since last month you've been asking several questions (here and on math.stackexchange) about solving $ay^2 = b + cx^4$ in various $p$-adic fields. It'd be nice to include context when asking such questions (related to computing Selmer groups?). | |
| Jul 5, 2023 at 17:17 | comment | added | KConrad | The number $17$ is a fourth power in $\mathbf Z_2$ (since $f(t) = t^4 - 17$ has $|f(3)|_2 < |f'(3)|_2^2$, so there is a root in $\mathbf Z_2$ by Hensel's lemma, and one root is $1+2+2^4 + 2^6 + 2^9 + \cdots$), so by scaling $x$ the equation can be simplified to $x^4 + y^2 = -1$ in extensions of $\mathbf Q_2$. | |
| Jul 5, 2023 at 16:34 | history | asked | Duality | CC BY-SA 4.0 |