# Bijective polynomials

Let $P\in\mathbb{R}[x,y]$ be called a bijective polynomial iff these two conditions hold.

• If $a,b\in\mathbb{Z}^+$, $P(a,b)\in\mathbb{Z}^+$.
• For each $x\in\mathbb{Z}^+$, a unique pair of integers $a,b\in\mathbb{Z}^+$ exists such that $P(a,b)=x$.

I then wish to find all bijective polynomials. My conjecture is that the polynomial $$P(x,y)=\frac{1}{2}(x^2+y^2+2xy-x-3y+2)$$ and its transpose are the only ones, and I've tried to prove that no such polynomials exist with degree greater than 2 via some kind of growth argument, but I haven't been able to do so.

Here's a (very crude) visualization of how the bijective polynomial given above works. • This polynomial is called the Cantor pairing function. – Arnaud Mortier Mar 25 '18 at 21:46
• @ArnaudMortier That's an open question? Has there been any significant progress? – Anonymous Pi Mar 25 '18 at 21:48

## 1 Answer

This is a century old conjecture. These two polynomials are known as the Cantor pairing polynomials.

Some results are known:

1. They are indeed the only two quadratic polynomials realising a bijection between $\Bbb Z^+$ and $\Bbb Z^+\times \Bbb Z^+$. [Fueter-Polya, 1923]

2. There are no such polynomials in degree 1, 3 or 4. [Lew-Rosenberg, 1978]

Reference: https://arxiv.org/pdf/1512.08261.pdf