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This came up in a discussion with a colleague of mine, who studies PDEs. He was asking for a function $f \colon \mathbb{N} \rightarrow \mathbb{N}$ such that, for all but finitely many $n$, the equation $$ x^4 + y^4 = n $$ has at most $f(n)$ solutions in integers $x,y$. My own feeling was that there should be a constant function $f$ that does the trick.

Indeed, if the "weak Lang conjecture" (see below) is true, then even the number of rational solutions to the above equation is bounded uniformly over all $n$. Assuming the weak Lang conjecture then, we can take $f$ to be a constant function. My question is whether we can prove this statement without assuming any unknown conjecture:

Does there exist $N \in \mathbb{N}$ such that, for all $n \in \mathbb{N}$, the equation $x^4+y^4=n$ has at most $N$ solutions in integers $x,y$?

The statement of the weak Lang conjecture is as follows: if $X$ is a variety of general type over a number field $K$, then the set $X(K)$ of rational points of $X$ is not Zariski dense. In their article "Uniformity of rational points" (JAMS, 1997), Caporaso, Harris, and Mazur prove that this conjecture implies the existence of constants $B(K,g)$ such that every smooth genus $g$ curve $X$ over a number field $K$ has at most $B(K,g)$ rational points (see Theorem 1.1 in their paper; note that they probably assume $X$ irreducible as well, although they do not explicitly state this).

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  • $\begingroup$ Your question is stated slightly confusingly - your friend asks for a bound depending on $n$ but you ask for a bound independent of $n$, and perhaps this is what confused ACL. $\endgroup$ Jan 27 '14 at 19:05
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    $\begingroup$ We certainly have $f(n)\ll_\epsilon n^\epsilon$ for any $\epsilon>0$, and we should have $f(n)\leq 16$ but this is widely open. See my comments to ACL's response. $\endgroup$
    – GH from MO
    Jan 27 '14 at 19:09
  • $\begingroup$ @GHfromMO what is the current record for number of solutions? $\endgroup$
    – joro
    Jan 28 '14 at 5:49
  • $\begingroup$ Euler noticed that $59^4+158^4=133^4+134^4$. This already gives $16$ solutions to $x^4+y^4=n$, where $n=635318657$ (apply the symmetries $(x,y)\mapsto(\pm x,\pm y)$ and $(x,y)\mapsto(y,x)$). According to Hardy and Wright then, this must be conjectured as being the record number of solutions for any $n$. $\endgroup$
    – RP_
    Jan 28 '14 at 11:17
  • $\begingroup$ @joro: I don't know what is the best current bound on $f(n)$. $\endgroup$
    – GH from MO
    Jan 28 '14 at 14:46
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Here are two papers that do not allow to conclude positively to your question. The reason is that one does not know how to bound uniformly the Mordell-Weil rank of abelian varieties of given dimension over a given number fields, even in well behaved families.

1. A general bound depending on the size of the coefficients

In his paper ``Borne polynomiale pour le nombre de points rationnels des courbes'', Journal de théorie des nombre de Bordeaux 23 no. 1 (2011), p. 251-255, Gaël Rémond has given a general explicit bound for the number of solutions of polynomial equations in two variables with coefficients in a number field, assuming the equation has finitely many solutions, of course. By Faltings's theorem, this is the case of your curve, so his bound applies and says that there are at most $n^{2^{3^{16}}}$ solutions.

NB. The exponent of $n$ is equal to $1.721783764\, 10^{12958354}$ and this bound is both unpractical and certainly non-optimal.

NB. Rémond bounds the Mordell-Weil rank in terms of the size of the equation.

2. A bound for twists of a given curve

The paper A uniform bound for rational points on twists of a given curve, J. London Math. Soc. (2) 47 (1993) 385-394, by Joseph Silverman shows a partial uniformity of the number of rational points among twists of a given curve. However, the bound depends on the Mordell-Weil rank of the specific curve, so is not really uniform.

He also gives the example of Catalan curves of the form $ax^m+bx^n=1$, where $m$ and $n$ are fixed and $a,b$ varies among non-zero rational numbers.

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    $\begingroup$ Wait, this bound depends on $n$, right? I am asking for a bound that is independent of $n$. $\endgroup$
    – RP_
    Jan 27 '14 at 16:13
  • $\begingroup$ Looking at the abstract, it seems that $3^{2^{3^{16}}}$ will do. $\endgroup$
    – Matt F.
    Jan 27 '14 at 16:55
  • $\begingroup$ @MattF. No, his $M$ has to be an upper bound for the coefficients. By the way, in a previous work, he had obtained the bound $\exp(5^{4^4}(\log (n))(\log\log(n)))$ which is better. (The first bound I copied is better in $n$, but worse in the degree.) $\endgroup$
    – ACL
    Jan 27 '14 at 17:16
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    $\begingroup$ An easy upper bound is $c_\epsilon n^\epsilon$ for any $\epsilon>0$, where $c_\epsilon$ is a suitable constant. In fact even the number of solutions of $x^2+y^2=n$ satisfies this bound. $\endgroup$
    – GH from MO
    Jan 27 '14 at 18:58
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    $\begingroup$ According to Hardy-Wright: An introduction to the theory of numbers, Chapter XXI, we should have $f(n)\leq 16$, but this is not known. $\endgroup$
    – GH from MO
    Jan 27 '14 at 19:01

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