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Consider the surface over the rationals $$ x^4+y^4=z^4+t^4 \qquad (1)$$

Consider parametrizations of form: $$ f_1(u)^4+f_2(u)^4=f_3(u)^4+f_4(u)^4 \qquad(2) $$ where $f_i$ are polynomials with integer coefficients, pairwise $f_i(u) \ne \pm f_j(u),i \ne j$ and $f_i$ coprime.

Let $d=\max \{\deg(f_i)\}$.

Q1 For fixed $d$, are there only finitely many parametrizations (2)?

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  • $\begingroup$ Dear joro, maybe this is a little more complicated than I thought. Let me delete my previous comment until I get things straight in my mind. (I still don't like the term "parametrizations" in this context, though.) $\endgroup$
    – Lazzaro Campeotti
    Commented Apr 20, 2016 at 13:21
  • $\begingroup$ @potentiallydense I don't mean complete parametrization, since it might make the surface rational. Do you suggest changing the wording or editing? Maybe "partial parametrization" is better? $\endgroup$
    – joro
    Commented Apr 20, 2016 at 13:29
  • $\begingroup$ I would say "parametrizations of rational curves". $\endgroup$
    – Lazzaro Campeotti
    Commented Apr 20, 2016 at 13:30
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    $\begingroup$ OK, let me contradict my previous comment (where I was only thinking about actual curves on the surface, not maps). Given one parametrization of a curve as in (2), don't you get infinitely many more with the same degree by making substitutions $u \mapsto au+b$ (for integers $a$ and $b$)? $\endgroup$
    – Lazzaro Campeotti
    Commented Apr 20, 2016 at 14:43
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    $\begingroup$ More generally you can replace $f_i(u)$ by $(\gamma u+\delta )^df_i(\dfrac{\alpha u+\beta }{\gamma u+\delta } )$. Up to these substitutions, the answer is positive. Indeed you are looking at rational curves of degree $\leq d$ on your surface; there is a finite number of families of curves of degree $\leq d$, and each family contains finitely many rational curves. $\endgroup$
    – abx
    Commented Apr 20, 2016 at 15:17

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