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The question Polynomial representing all nonnegative integersPolynomial representing all nonnegative integers asked by Poonen has been on MO for a very long time. (It should possibly be mentioend that it was asked elsewhere earlier, in particular on Poonen's website.) It received various interesting answers, but none of them solved the problem.

The question is short so I reproduce it:

Is there a 2-variable polynomial $$f(x,y) \in \mathbf{Q}[x,y]$$ such that $$f(\mathbf{Z} \times \mathbf{Z})=\mathbf{N}$$?

For $$3$$ (and more variables) it is well known such a polynomial exists (basically going back to Gauss).

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pageman
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The question Polynomial representing all nonnegative integers asked by Poonen ishas been on MO sincefor a very long time. (It should possibly be mentioend that it was asked elsewhere earlier, in particular on Poonen's website.) It received various interesting answers, but none of them solved the problem.

The question is short so I reproduce it:

Is there a 2-variable polynomial $$f(x,y) \in \mathbf{Q}[x,y]$$ such that $$f(\mathbf{Z} \times \mathbf{Z})=\mathbf{N}$$?

For $$3$$ (and more variables) it is well known such a polynomial exists (basically going back to Gauss).

user9072

The question Polynomial representing all nonnegative integers asked by Poonen is on MO since a very long time. (It should possibly be mentioend that it was asked elsewhere earlier, in particular on Poonen's website.) It received various interesting answers, but none of them solved the problem.

The question is short so I reproduce it:

Is there a 2-variable polynomial $$f(x,y) \in \mathbf{Q}[x,y]$$ such that $$f(\mathbf{Z} \times \mathbf{Z})=\mathbf{N}$$?

For $$3$$ (and more variables) it is well known such a polynomial exists (basically going back to Gauss).

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