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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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Do you have representations for 2 and for 3? I didn't find any with $1\le x\le y\le1000$. – Gerry Myerson Jul 17 '14 at 1:16
@Gerry: there might be something wrong with your code (maybe you checked $x<y$). $(4,4,5)$ for $k=3$, and $(1,1,0)$ for $2$. – Zack Wolske Jul 17 '14 at 1:26
@Zack, $4^4+4^4=256+256=512\ne128=125+3=5^3+3$. You must be looking at cubes. But I accept the correction as regards $k=2$. – Gerry Myerson Jul 17 '14 at 1:38
@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. – Gerry Myerson Jul 17 '14 at 2:25
$x^4 + y^4 = z^3 w + k w^4$ 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 $x^4 + y^4 + z^4 = w^4$. – Will Sawin Jul 23 '14 at 1:51

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