# How to express birational equivalence of Diophantine equation $x^4+y^4=z^2$ and elliptic curve?

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 $$x^4$$?

• 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... Commented Jul 22, 2020 at 0:29
• @AlexeyBurdin Ok, so it's essentially a free variable? That would make sense. How would I write that like I did for $(x, y, z)$? Commented Jul 22, 2020 at 0:31

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).