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I think $x^4+ux^2y^2+y^4=z^2$ is an elliptic curve, so how should I transform it into Weierstrass form? Either by hand or by software like MAGMA is fine. I am new to MAGMA and I tried something similar to this, but it seems like the "Curve" function requires the function to be homogeneous, which I don't know why and how to get around.

Edit: Here $u$ is a parameter, and I meant to ask that for every $u$, what is the corresponding Weierstrass form, with $u$ as a parameter in it. The base field considered is $\mathbb{Q}$.

fp1
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    Is that an equation in a weighted projective space? Because it's not homogeneous otherwise. Also, there's a $2$ in the title and a $u$ in the post. What is the base field you are interested in? A number field/finite field or a function field $K(u)$? It would be helpful to add these details. – Arkady Jul 11 '22 at 21:08
  • As @Arkady, comments, there might be a typo or misunderstanding: as it stands, this single equation in 3 variables is not homogeneous, so does not obviously define a projective curve. Should the $z^2$ be a $z^4$, for example? – paul garrett Jul 11 '22 at 21:13
  • The equation $x^4+ux^2y^2+y^4=z^2$ is of a surface and not a curve, unless there is a typo in the exponent of $z$ and considered projectively. – Somos Jul 11 '22 at 21:27
  • @Arkady Thank you for pointing out the typo in the title. It should be $u$, which is a parameter. I have also tried to make this clearer by adding an explanation. The power of $z$ however is intended to be 2, not 4. The case for $u=\pm1$ is mentioned here: https://math.stackexchange.com/a/1764859/765986 Except that it is a 2-isogeny in the answer. Is that the best transform one can do in this case? – fp1 Jul 13 '22 at 02:44
  • @Somos The question partially comes from an answer I linked in the comment above. I was also slightly confused when I saw this answer. I wonder if one can do more than just a 2-isogeny basically. – fp1 Jul 13 '22 at 02:48
  • @paulgarrett The power of $z$ is intended to be 4. This confusion/question originated from an answer I linked above in the comments, and I wonder if this 2-isogeny is the best one can do? – fp1 Jul 13 '22 at 02:50
  • I'm still not understanding the context... can you be more "effusive" in your describing the issue you're addressing? As in other comments, the $z^2$, as opposed to $z^4$, creates problems in the very posing of your question. A more verbose framing of your issue would surely help people help you... :) – paul garrett Jul 13 '22 at 02:59
  • @paulgarrett Sure. So in this link https://math.stackexchange.com/a/1764859/765986 Prof. Elkies's answer says $x^4\pm x^2y^2+y^4=z^2$ are elliptic curves, and gave a 2-isogeny to map them to a Weierstrass froms because a 2-isogeny was good enough there. Then, during a conversation with another professor, he also mentioned that $x^4+ux^2y^2+y^4=z^2$ parametrizes elliptic curves, but he couldn't remember references for this. So now I am assuming indeed $x^4+ux^2y^2+y^4=z^2$ is an elliptic curve. I am also confused by the $z^2$, but I think the equation is projective, so no need for homogeneous? – fp1 Jul 13 '22 at 03:16
  • Maybe $z$ is "just" a parameter? – paul garrett Jul 13 '22 at 03:27
  • @paulgarrett Prof. Elkies is working in weighted projective space. See example 1.2 here: https://homepages.warwick.ac.uk/~masda/surf/more/grad.pdf – Arkady Jul 13 '22 at 03:30
  • Thanks, @Arkady! :) – paul garrett Jul 13 '22 at 16:48
  • I supplied the missing details of Elkies in my answer. – Somos Jul 13 '22 at 20:30
  • Two obvious families of solutions have $u=\pm{2}$ and $z=x^2 \pm y^2$. – Geoffrey Trang Aug 01 '22 at 02:18

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The good answer by Elkies is missing a little detail which I now supply and explain.

Suppose that $\,u,x,y,z\,$ are integers with $\,y\ne0.\,$ Given the equation $$ x^4 + ux^2y^2 + y^4 = z^2, $$ Elkies suggests to define $\,t = x/y\,$ but omits that he also assumes $\,v=z/y^2.\,$ Thus $$ 1 + u t^2 + t^4 - v^2 = (x^4 + u x^2 y^2 + y^4-z^2)/y^4. $$

Similarly, use $\,X = x^2/y^2,\; Y = xz/y^3\,$ to get the Weierstrass form $$ X^3 \!+\! uX^2 \!+\! X \!-\! Y^2 \!=\! (x^4 \!+\! ux^2y^2 \!+\! y^4 \!-\! z^2)x^2/y^6. $$

Thus, rational solutions to $\, X^3 + uX^2 + X = Y^2 \,$ (assuming $\,X = t^2$) correspond to integer solutions to $\, x^4 + ux^2y^2 + y^4 = z^2 \,$ by using $\,x = y\,t,\; z = Yy^2/t.$

For $\,u=-1\,$ the elliptic curve with LMFDB label 24.a5 $\,y^2 = x^3-x^2+x\,$ has only three rational points which are $\,(0,0),\;(1,1),\;(1,-1)\,$ and also the point at infinity.

For $\,u=1\,$ the elliptic curve with LMFDB label 48.a5 $\,y^2 = x^3+x^2+x\,$ has only one rational point which is $\,(0,0)\,$ and also the point at infinity.

Somos
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