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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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@Harry: I'm not sure if I'm understanding your statement about multivariate versus multivariable; however, if you are saying that the term multivariate polynomial doesn't mean anything, I'd have to say that I disagree. mathworld.wolfram.com/MultivariatePolynomial.html –  MLevi Dec 9 '09 at 6:29
    
Multivariate usually has something to do with statistics, in my experience, although my experience consists of me looking it up on wikipedia an hour ago. The terminology doesn't sit right with me. It sounds artificial and bad. –  Harry Gindi Dec 9 '09 at 7:05

2 Answers 2

up vote 5 down vote accepted

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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(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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