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The background to the question:

$$a^4+b^4=c^4+d^4 \label{1}\tag 1 $$

Tito Piezas, Tomita & others have recently given some parametric solutions on Math stack exchange & Math overflow. In math literature there are parametric solutions given in Dickson’s book (vol 2), which includes solutions by Euler & others. Solutions are shown by Zajta (In a AMS journal paper). For degree two we have general solution for $a^2+b^2=c^2+d^2.$ For degree three $a^3+b^3=c^3+d^3$ two general solution has been given. One by Noam Elkies & second by Ajai Choudhry. Even though this problem of the quartic equation (# \ref{1} above) has been around since the time of Euler (for over 200) years a general solution has been evasive. Eight relevant Links are shown below:

Remark:

There are more than a couple of dozen parametric solutions available for equation \eqref{1}. There is a possibility that one of them could be a general solution or maybe not. There are two options. First is, someone needs to write an algorithm to see if the (1420 different) numerical solutions (link as shown above & given at Jaroslaw Wrobelewski website -uni.wroc) are satisfied by one of the published parametric solution. The other option is that someone needs to give a general solution along with a proof. Any response to the above will be appreciated.

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    $\begingroup$ This degree 4 equation defines a K3 surface, which is not unirational (in contrast to the degree 3 case, where it is unirational). Basically by definition of "unirational," this means there's not a nice formula for the general solution of the type you are probably imagining. $\endgroup$
    – sdr
    Commented Oct 19, 2023 at 18:06
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    $\begingroup$ @sdr, I think you should make your comment an answer, perhaps with some further details! :) $\endgroup$
    – paul garrett
    Commented Oct 19, 2023 at 19:12
  • $\begingroup$ "To add to Silverman's remark: In the case (x1)^d+(x2)^d=(x3)^d+(x4)^d, with d≥5, depending on one's viewpoint a "formal for a general solution" should indeed exist" $\endgroup$
    – David
    Commented Oct 26, 2023 at 15:28
  • $\begingroup$ @Daniel Longhran, Portion of your previous comment is pasted here [ "To add to Silverman's remark: In the case, (x1)^d+(x2)^d=(x3)^d+(x4)^d, with d≥5, depending on one's viewpoint a "formal for a general solution" should indeed exist" ]. So what is view on exponent (d=4). With hindsight from the comments of Naom Elkies & Joe Silverman I think viewers would appreciate if they see a (part two) of your answer. $\endgroup$
    – David
    Commented Oct 26, 2023 at 15:42
  • $\begingroup$ @David: I dont understand what you are asking. Can you please be more precise? If you have a separate question it may be good to ask a new one $\endgroup$
    – Daniel Loughran
    Commented Oct 28, 2023 at 14:30

1 Answer 1

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This question has a false premise; there is no such thing as a "formula for a general solution" in this case, exactly for the reason that user sdr describes.

I interpret the question as follows. The OP is looking for polynomials $f_1(x_0,\dots,x_n), \dots, f_4(x_0,\dots,x_n)$ such that any solution to the equation can be written as $$a = f_1(x_1,\dots,x_n), \dots, d = f_4(x_1,\dots,x_n)$$ for some choice of $x_i \in \mathbb{Q}$. Such a formula exactly corresponds to a rational map of varieties $$\mathbb{A}^n \dashrightarrow S$$ over $\mathbb{Q}$, where $S$ denotes the corresponding surface in $\mathbb{P}^3$. However there is no such map even over $\mathbb{C}$. This is because $S$ defines a K3 surface and K3 surfaces are not unirational.

This is all explained on the wikipedia page for K3 surfaces (here it is explained that they are not uniruled, which implies they are not unirational).

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    $\begingroup$ (Corrected following @Joe Silverman's observtions...) Exactly. This is a fancier version of a point that Joe Lipman made to me many years ago, that there are no (non-constant) rational-function solutions to Fermat equations $x^n+y^n=1$ for $n>2$, because those curves are not genus zero. It was dumbfounding to me at the time! :) That is, in the face of extensive computations I'd done (oof), "it is obvious". :) That is, maybe concepts are worth something!?! :) $\endgroup$
    – paul garrett
    Commented Oct 20, 2023 at 15:56
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    $\begingroup$ @paulgarrett Right. And moving in the opposite direction (dimension-wise), an analogous statement holds for $$x_1^n+x_2^n+\cdots+x_d^n=1$$ whenever $n\ge d+1$. This is because the canonical divisor of this smooth hypersurface in $\mathbb P^d$ is $\mathcal O(n-d-1)$, so its anti-canonical divisor $\mathcal O(d+1-n)$ is not ample if $n\ge d+1$; but the anti-canonical divisor of a rational variety is ample. As you say, hooray for general theory! $\endgroup$
    – Joe Silverman
    Commented Oct 20, 2023 at 20:00
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    $\begingroup$ @paulgarrett I'm usually hesitant about posting an MO "answer" that doesn't actually answer the question. So I tend to think that generalizations of this sort belong in the comments section. $\endgroup$
    – Joe Silverman
    Commented Oct 20, 2023 at 22:19
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    $\begingroup$ @JoeSilverman Your comment does answer the question, because the question concerns the special case $(n,d) = (4,3)$ (once we work over a field in which $-1$ is a 4th power). $\endgroup$
    – Noam D. Elkies
    Commented Oct 22, 2023 at 3:10
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    $\begingroup$ To add to Silverman's remark: In the case $x_0^d + x_1^d = x_2^d + x_3^d$ with $d \geq 5$, depending on one's viewpoint a "formal for a general solution" should indeed exist. This is because the Bombieri-Lang conjecture predicts that all but finitely many rational points lie on finitely many rational and elliptic curves. So it just suffices to list these curves with their rational points and the rational points not on the curves. However there is no hope of proving anything like this at all with current tools! $\endgroup$
    – Daniel Loughran
    Commented Oct 24, 2023 at 8:49

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