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Let $E:y^2=x^3-17$ be an elliptic curve.

It is known that rank$(E/\Bbb{Q})=0$.

(For example, prop $6.5$, $362$p in Silverman's book 'The arithmetic of elliptic curves')

Over $K=\Bbb{Q}(i)$, what is the rank of $E$ ?

I couldn't find this curve ($E/\Bbb{Q}(i)$) in LMFDB (I don't know what is known about the rank over number fields except for $\Bbb{Q}$)

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    $\begingroup$ $r(E/\mathbb{Q}(i)) = r(E/\mathbb{Q}) + r(E^{(-1)}/\mathbb{Q}) = 0 + 2$ with the quadratic twist $E^{(-1)}$ by $-1$. $\endgroup$
    – user471019
    Commented Apr 30, 2023 at 15:25
  • $\begingroup$ Could you tell me why your $=$ holds and $r(E^{(-1)}//\Bbb{Q})=2$? (Can you apply the same way to another quadratic field, for example, $\Bbb{Q}(\sqrt{2})$? $\endgroup$
    – Duality
    Commented Apr 30, 2023 at 16:50
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    $\begingroup$ $=$ is a general statement about quadratic twists, [Silverman 2nd ed, exercise 10.16]. I computed the rank of the twisted curve using Magma: Rank(QuadraticTwist(E,-1)); $\endgroup$
    – user471019
    Commented Apr 30, 2023 at 17:13
  • $\begingroup$ Thank you very much. Let $D \in \Bbb{Q}^{\times}$, $E_D$ be quadratic twist of $E/\Bbb{Q}$. I also used Magma to calculate rank of $E_D(\Bbb{Q})$ for some $D$, and the result is always $2$. Possibly we can say rank of quadratic twist is always $2$ in this elliptic curve ? $\endgroup$
    – Duality
    Commented May 1, 2023 at 3:41
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    $\begingroup$ No, the rank of the quadratic twist by $-47$ is $1$. I found it using HeegnerDiscriminants(E,1,100); $\endgroup$
    – user471019
    Commented May 1, 2023 at 5:58

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