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To be precise, I am asking:

Does there exist an integer $k$ such that there do not exist (possibly negative) integers $x,y,z$ satisfying $x^4+y^4=z^3+k$?

Heuristically the answer must be yes, in fact, one expects that almost every $k$ should work (this is just because the sum of the reciprocals of the exponents is $\frac13+\frac14+\frac14 = \frac56<1$, so one ought to expect that up to a large bound $N$, something like $N^{5/6}$ values of $k$ are representable). For all I know, $-2$ and $4$ might already be two examples of numbers which can't be represented as sums of two fourth powers minus a cube, but I haven't been able to prove this.

What I can prove is that there are no local obstructions: for any integers $n,k$ with $n\ne 0$, we can find integers $x,y,z$ with $(x,y,z)=1$ such that $x^4+y^4\equiv z^3+k\pmod{n}.$ Using the Chinese Remainder Theorem and Hensel's Lemma one can quickly reduce this claim to the case that $n=p$ is a prime. The most interesting case is when $p\equiv 1\pmod{12}$ and $p\nmid k$, and in this case we can use a trick similar to the proof of Chevalley-Warning to count the number of solutions $N_p$ modulo $p$. We start with the easy congruence

$N_p \equiv \sum_{x,y,z} (1-(x^4+y^4-z^3-k)^{p-1}) \equiv -\sum_{x,y,z}\sum_{a+b+c+d=p-1} \frac{(p-1)!}{a!b!c!d!} x^{4a}y^{4b}(-z)^{3c}(-k)^d\pmod{p},$

and then note that if we fix $(a,b,c,d)$ and sum over $x,y,z$, we can only get a nonzero contribution when $p-1\mid 4a,4b,3c$ and $a,b,c>0$. From this we see that

$N_p \equiv \frac{(p-1)!}{\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{3}\right)!\left(\frac{p-1}{6}\right)!} (-k)^{\frac{p-1}{6}}\not\equiv 0\pmod{p},$

so in particular $N_p \ne 0$.

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  • $\begingroup$ Do you have representations for 2 and for 3? I didn't find any with $1\le x\le y\le1000$. $\endgroup$
    – Gerry Myerson
    Commented Jul 17, 2014 at 1:16
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    $\begingroup$ @Zack, OK, I see why I missed those: I was taking $z$ to be the floor of $\root3\of{x^4+y^4}$, which misses some cases with $x,y$ very small. $\endgroup$
    – Gerry Myerson
    Commented Jul 17, 2014 at 2:25
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    $\begingroup$ $x^4 + y^4 = z^3 w + k w^4$ is a smooth K3 surface. So one could try to understand the geometry of this surface and gain insight into at least the rational points by e.g. finding curves on it, similar to Elkies' work on $x^4 + y^4 + z^4 = w^4$. $\endgroup$
    – Will Sawin
    Commented Jul 23, 2014 at 1:51
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    $\begingroup$ For $k=-38$, the smallest solution appears to be $365^4+103^4=2614^3-38$. What are some other $k$ for which a solution exists but the smallest solution is quite "big"? $\endgroup$
    – Jeppe Stig Nielsen
    Commented Oct 6, 2016 at 14:11
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    $\begingroup$ @JeppeStigNielsen: I found these three: $1680^4+ 3466^4- 53401^3= 135$, $9637^4+10103^4- 267044^3= -942$ and $14939^4+16063^4- 488233^3= 1465$. Not too exciting. $\endgroup$
    – Yaakov Baruch
    Commented Aug 10, 2017 at 20:35

1 Answer 1

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Section 5 of my paper http://maths.nju.edu.cn/~zwsun/179b.pdf is closely related to your question. You can find there a theorem, two conjectures and some heurisitic arguments with explanations. The heuristic arguments (not rigorous) there suggest that there are integers not of the form $x^4+y^4-z^3$ with $x,y,z$ positive integers. In view of this, I believe that there are infnitely many integers $k$ which cannot be written as $x^4+y^4-z^3$ with $x,y,z$ integers.

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  • $\begingroup$ But there are definitely integers that can be expressed as $x^4 + y^4 -z^3 = k$, most probably the smallest integer being $9 = 2^4 + 1^4 - 2^3$. The other is $633 = 2^4 + 5^4 - 2^3$. $\endgroup$
    – user25406
    Commented Sep 3, 2023 at 17:39

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