I have a series of change of variables to go from the Diophantine equation $x^4 + y^4 = z^2$ to the elliptic curve $y^2 = x^3 - 4x$ that is supposedly a bijection (bar a finite number of trivial solutions):

$$ \begin{align} x^4+y^4=z^2 \\ v^2 = u^4+1 && (u, v) &= (y/x, z/x^2) \\ r^2 + 2rs^2 = 1 && (r, s) &= (v-u^2, u) \\ a^3 + 2b^2 = a && (a, b) &= (r, rs) \\ y_1^2 = x_1^3 - 4x_1 && (x_1, y_1) &= (-2a, 4b) \end{align} $$

My question is, if I'm going in the inverse direction, how do I find $x$ in terms of $x_1, y_1$ if I only have two variables in the elliptic curve? Further, how could I write this into one singular change of variables? I can write the forward change as one:

$$ (x, y, z) \rightarrow \left(-2 \frac{z-y^2}{x^2}, 4 \frac yx \left(\frac{z-y^2}{x^2}\right)\right) $$

For the reverse I can make it up to $v^2 = u^4 + 1$ with the map:

$$ (x_1, y_1) \rightarrow \left(-\frac{y_1^2-2x_1^3}{4x_1^2}, -\frac{y_1}{2x_1}\right) $$

How would I notate going from the 2nd equation to the first, would $x$ just be a free variable and I multiply each side of $v^2=u^4+1$ by $x^4$?

  • $\begingroup$ When you go back from 2nd to 1st equation you need to multiply by some $x^4$ to make the things integer back from rationals (so maybe taking a multiple of the [common?] denominator as $x$ will do). If this makes sense... $\endgroup$ Commented Jul 22, 2020 at 0:29
  • $\begingroup$ @AlexeyBurdin Ok, so it's essentially a free variable? That would make sense. How would I write that like I did for $(x, y, z)$? $\endgroup$
    – Andrew Li
    Commented Jul 22, 2020 at 0:31

1 Answer 1


I am pretty late to the party, but hopefully this helps. So you have a series of maps, to show that they are birational (at least from the second to the last - the first is not a bijection if $x,y,z \in \mathbb{Q}$ instead of $\mathbb{Z}$) it suffices to show it at each stage.

As you show, $x,y,z \in \mathbb{Z}$ satisfying $x^4 + y^4 = z^2$ yield $u = y/x$ and $v = z/x^2$ such that $v^2 = u^4 + 1$. Conversely given such $u, v$ we construct $x,y,z$ by clearing denominators - check that (with appropriate coprimality conditions) this is a bijection.

To check the map $(u, v) \to (v - u^2, u)$ is birational note that $(r, s) \to (s, r + s^2)$ goes the other way.

Similarly to check the map $(r,s) \to (r, rs)$ is birational note that $(a,b) \to (a, b/a)$ works when $a \neq 0$ (i.e., a finite number of points).

The final coordinate change is clearly birational.


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