It is known that the polynomial $f(n,m)=\frac{1}{2}(n+m)(n+m+1)+m$ defines bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ (Put pairs of $\mathbb{N}$ into the semi-infinite matrix and count them by diagonals). Does there exist a polynomial bijection $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$? The question is related to the open question about polynomial bijection $\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$ here.
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5$\begingroup$ Related but not decisive: thehcmr.org/issue1_2/bert_and_ernie.pdf $\endgroup$– Benjamin DickmanCommented Dec 28, 2012 at 6:53
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6$\begingroup$ [It's also AMM 6028, which remains unsolved as far as I know.] $\endgroup$– Benjamin DickmanCommented Dec 28, 2012 at 7:02
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4$\begingroup$ @Dickman: It took a me a while to know what you were talking about. Let me add the link books.google.es/books?id=KX6D6hefyA0C&pg=217 $\endgroup$– boumolCommented Dec 28, 2012 at 11:44
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6$\begingroup$ @Dicman and @Boumol: Thank you for interesting references. Interesting, AMM6028 asks for polynomials with integer coefficients. In fact, the bijection $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ I know has rational coefficients. Does there exist polynomial $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ bijection with integer coefficients? $\endgroup$– Lev GlebskyCommented Dec 28, 2012 at 18:32
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3$\begingroup$ Updated: zacharyabel.com/papers/Favorite-Problem_A07_HCMR.pdf $\endgroup$– Steve DCommented Mar 9, 2016 at 22:09
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1 Answer
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It is an open problem. Maximal results about bijections from $\mathbb N\times \mathbb N$, $\mathbb Z\times \mathbb N$, $\mathbb Z\times \mathbb Z$ to $\mathbb N$ are contained in
John S. Lew, Arnold L. Rosenberg, Polynomial indexing of integer lattice-points I. General concepts and quadratic polynomials, J. Number Theory 10 (1978) pp 192-214, doi:10.1016/0022-314X(78)90035-5.
Polynomial indexing of integer lattice-points II. Nonexistence results for higher-degree polynomials, J. Number Theory 10 (1978) pp 215-243, doi:10.1016/0022-314X(78)90036-7
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1$\begingroup$ I edited the references and links. $\endgroup$– David Roberts ♦Commented Mar 10, 2016 at 7:05