You do get zero density for $x^k - y^k$ with, say, $x,y > 0,$ as $ x^k - y^k = (x-y) (x^{k-1} + \cdots + y^{k-1} ) $ and either $x=y$ or $| x - y| \geq 1,$ so the number of $(x,y)$ pairs with $0 < x^k - y^k \leq n$ is no larger than $n^{2/(k-1)}.$ As soon as $k \geq 4$ we get zero density.

However, it is now suspected that $\pm x^3 \pm y^3 \pm z^3$ gives full density, that being 7/9. Also, by easy identities, all numbers are the mixed sum of five cubes, so $v(3) \leq 5$

I would switch to asking your question with three variables instead of five, just completely ignore congruences and focus on your zero density, which seems a clever idea to me...

You do get zero density for $x^k - y^k$ with, say, $x,y > 0,$ as $ x^k - y^k = (x-y) (x^{k-1} + \cdots + y^{k-1} ) $ and either $x=y$ or $| x - y| \geq 1,$ so the number of $(x,y)$ pairs with $0 < x^k - y^k \leq n$ is no larger than $n^{2/(k-1)}.$ As soon as $k \geq 4$ we get zero density. Meanwhile, $x^3 - y^3$ also gives zero density, but this relates to the set of numbers up to $n$ that are represented by the positive quadratic form $x^2 + x y + y^2$ and the count is constant times $\frac{n}{\sqrt {\log n}}.$

However, it is now suspected that $\pm x^3 \pm y^3 \pm z^3$ gives full density, that being 7/9. Also, by easy identities, all numbers are the mixed sum of five cubes, so $v(3) \leq 5$

I would switch to asking your question with three variables instead of five, just completely ignore congruences and focus on your zero density, which seems a clever idea to me...

You do get zero density for $x^k - y^k$ with, say, $x,y > 0,$ as $ x^k - y^k = (x-y) (x^{k-1} + \cdots + y^{k-1} ) $ and either $x=y$ or $| x - y| \geq 1,$ so the number of $(x,y)$ pairs with $0 < x^k - y^k \leq n$ is no larger than $n^{2/(k-1)}.$ As soon as $k \geq 4$ we get zero density.

However, it is now suspected that $\pm x^3 \pm y^3 \pm z^3$ gives full density, that being 7/9. Also, by easy identities, all numbers are the mixed sum of five cubes, so $v(3) \leq 5$

I would start by asking your question with three variables instead of five...

You do get zero density for $x^k - y^k$ with, say, $x,y > 0,$ as $ x^k - y^k = (x-y) (x^{k-1} + \cdots + y^{k-1} ) $ and either $x=y$ or $| x - y| \geq 1,$ so the number of $(x,y)$ pairs with $0 < x^k - y^k \leq n$ is no larger than $n^{2/(k-1)}.$ As soon as $k \geq 4$ we get zero density.

However, it is now suspected that $\pm x^3 \pm y^3 \pm z^3$ gives full density, that being 7/9. Also, by easy identities, all numbers are the mixed sum of five cubes, so $v(3) \leq 5$

I would switch to asking your question with three variables instead of five, just completely ignore congruences and focus on your zero density, which seems a clever idea to me...