No, every Mordell curve has only finitely many integral points.

The footnote on page 6 here may be of interest: statweb.stanford.edu/~jtaylo/courses/stats352/notes/…

I don't think it applies to the OP's function for all $p > 0$, wolframalpha.com/input/…

In concrete examples, it is easy to disprove that a kernel of the form $c(x, y) = f(|x-y|)$ is positive definite, by computing the Fourier transform of $f$ up to some accuracy and looking for negative values.

@EmilJeřábeksupportsMonica I'm not confident enough to make a precise statement; the complexity of the Euclidean algorithm part will depend on the denominators and numerators of $AB$.

I want a left inverse because that allows me to cancel the $A$ in the proof of the equivalence: $BAw = w$. It just so happens that $AB$ appears in the end.

Does this help? If $E$ is given as $\{ v + Aw : w \in \mathbb Q^m \}$, $A \in M_{n, m}(\mathbb Q)$, $m \leq n$, and $B$ is a left-inverse for $A$, then $E \cap \mathbb Z^n \neq \varnothing$ is equivalent to $(1-AB) v \in \mathbb Z^n + A \mathbb Z^m$. We can find a basis of this full rank $\mathbb Z$-module and check if the coordinates of $(1-AB) v$ in that basis are integers.

I think the paper is: Elijah Liflyand, Necessary Conditions for Integrability of the Fourier Transform, Georgian Mathematical Journal, Volume 16 (2009), Number 3, 553–559

@user44191 how do you show that $f(A, A) \subset A$ for $A$ in the image of $f$?

Do you know what can be said about smooth $G$-equivariant maps $\mathbb R^n \to \mathbb R^m$, given representations of $G$ on both spaces? (motivation)