2
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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 $\displaystyle F(\bigoplus_{j=1}^{k} \mathbb{Z}) = \bigcup_{j=1}^{n(k)} f_{j}(\mathbb{Z})$ ? Think this question is really cool. What do you guys think? |
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5
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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. |
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4
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(This answer was written before fedja's appeared, and has been edited to incorporate one of his observations.) No. In particular, I claim that the image of $F(x, y) = x^2 + y^2$ is not of this form. Suppose there exist integer polynomials $f_1, ... f_n$ with the desired property. Since the image of an odd degree polynomial contains negative numbers, the polynomials $f_i$ must have degree at least $2$. However, $\displaystyle \sum_{i=1}^{n} \sum_{x \in \mathbb{Z}} \frac{1}{|f_i(x) + 1|}$ converges, but $\displaystyle \sum_{x, y \in \mathbb{Z}} \frac{1}{x^2 + y^2 + 1}$ diverges (since, for example, it contains a subsequence which is essentially the sum of the reciprocals of the primes congruent to $1 \bmod 4$). |
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