MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters.

share|cite|improve this question
There have been a number of questions related to this, including one of the highest-voted ones by Bjorn Poonen. You might search through existing questions. – Will Jagy Nov 7 '10 at 15:59
In particular, the first link tells us that this question is an open problem. – Martin Brandenburg Nov 7 '10 at 16:15
@Martin actually that is only asking about surjectivity when the domain is $\mathbb{Z}\times\mathbb{Z}$, but I agree that it has some bearing here. – David Roberts Nov 8 '10 at 2:49

Describing such bijections is an open problem. Maximal result (there is no other bijections among polynomials of degree not higher than 4) 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

share|cite|improve this answer
What, exactly, are you claiming to be an open problem? As far as I can see, the question asks for a list of examples. – Emil Jeřábek Mar 12 at 9:45
Ok, Describing such bijections is an open problem. – Mikhail Ivanov Mar 12 at 16:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.