Given the simultaneous Diophantine equations,


$$x^4+y^4 = (u^6+v^6)t^2\tag{2}$$

the only solutions seem to be for the first Pythagorean triple $u,v,w = 3,4,5$ which yield the elliptic curve,

$$x^4+y^4 = 193z^2$$

with initial solution $x,y = 18,31$. Or alternatively,

$$s^4+1 = 193t^2$$

which has initial rational point $s = 18/31$ (and an infinite more) and is birationally equivalent to an elliptic curve in Weierstrass form.

Question: Are there other co-prime $u,v$, such that (2) is solvable? (I did a quick search using Mathematica and didn't find any, though my search radius for $x,y$ could just have been too small.)

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    $\begingroup$ $x^4+y^4=193z^2$ is not an elliptic curve. $\endgroup$ – Álvaro Lozano-Robledo Apr 20 '13 at 15:02
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    $\begingroup$ @exploringnet: yes, i deleted that question as I forgot the requirement that $u,v,w$ be a Pythagorean triple (eq.1). Without it, eq. 2 has a lot of solutions. When I couldn't find others with eq.1 as a constraint, I decided to restore the question. $\endgroup$ – Tito Piezas III Apr 20 '13 at 15:12
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    $\begingroup$ The problem is not the degree, for instance, see Cassels, Lectures on elliptic curves, chapter 8. $\endgroup$ – Myself Apr 20 '13 at 15:13
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    $\begingroup$ Page 35. (Note that x^4+y^4 = z^2 is non-homogeneous with 3 variables in it.) $\endgroup$ – Myself Apr 20 '13 at 15:26
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    $\begingroup$ The point is that if it is a curve of genus 1 (with base point) and if you can bring it into its canonical form y^2 = x^3 + ax + b with a birational transformation then you can use standard methods to find points on it. I claim no further expertise, but this curve looks more like a surface to me. $\endgroup$ – Myself Apr 20 '13 at 15:36

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