Let $k$ be a positive integer greater than $1$ and suppose that $F \in \mathbb{Z}[x\_{1}, \ldots, x\_{k}]$.


Can we always find a natural number $n(k)$ and $f\_{1}, \ldots f\_{n(k)} \in \mathbb{Z}[x]$ such that


$\displaystyle F\Big(\bigoplus\_{j=1}^{k} \mathbb{Z}\Big) = \bigcup\_{j=1}^{n(k)} f\_{j}(\mathbb{Z})$ ?


Think this question is really cool. What do you guys think?



Decomposition result for multivariate polynomial
Not really. Take $F(x,y,z,t)=x^2+y^2+z^2+t^2$. Then the image consists of all non-negative integers (Lagrange 4-square theorem). On the other hand, any linear polynomial will give you negatives in the image and every polynomial of degree 2 and higher will give you a zero density set.


