Is There An Injective Cubic Polynomial Z2→Z\mathbb Z^2 \rightarrow \mathbb Z?

Earlier, I was curious about whether a polynomial mapping Z2Z could be injective, and if so, what the minimum degree of such a polynomial could be. I’ve managed to construct such a quartic and rule out the existence of such a quadratic, but this leaves open the question of whether a cubic might exist. Equivalently my question is:

Can a cubic polynomial of two variables with integer coefficients be injective?

My intuition is that there probably is such a function since there is a quadratic bijection N2N so if we allow ourselves an extra “degree” to compensate for the transition from N to Z, it seems like it ought to be sufficient. However, I have yet to come up with an example that I suspect of being injective nor any general method I might use to try to prove injectivity.

The Part of This Post That Isn’t A Question, But That Helps Motivate It Or Maybe Inspire Someone:

So far I have determined that there is an injective quartic and there is not an injective quadratic. To construct the quartic which is injective, note that the map f:N2N defined by f(x,y)=(x+y)2+y is injective and so is the map g:ZN defined by g(n)=2n2n. Then, one can set
as an injective polynomial (of degree 4) in the two variables.

No quadratic polynomial may exist because any integer valued polynomial of degree two has a (non-zero) multiple expressible as:
where P1 and P2 are polynomials with integer coefficients. Then, if we choose some y1 and a y2 such that y_1\equiv y_2 \pmod{4aP_1(y_1)}, we clearly have that P_1(y_1)\equiv P_1(y_2)\pmod a and that P_2(y_1)-P_2(y_2)=4aP_1(y_1)k for integer k. Then, we can choose two integers c_1 and c_2 such that c_1^2-c_2^2=P_2(y_2) – P_2(y_1)
c_1\equiv c_2 \equiv P_1(y_1)\equiv P_1(y_2) \pmod a
In particular the values
satify the above. Then, choosing x_1=\frac{c_1-P_1(y_1)}{a} x_2=\frac{c_2-P_1(y_2)}{a} yields that P(x_1,y_1)=P(x_2,y_2), since their difference is (c_1^2-c_2^2) – (P_2(y_2)-P_2(y_1)) which we chose the c‘s to make 0.

I have no idea how to approach the cubic case.

A Moderately Surprising Computational Result

Using Mathematica, I determined that there is no polynomial of degree three with integer coefficients with absolute value 2 or less which is injective over the domain (\mathbb Z \cap [-2,2])^2. This surprises me, but it such a small set of polynomials that it might not mean anything other than that we might expect large-ish coefficients if a suitable polynomial does exist. (It could also be indicative of the fact that no such polynomial exists). I would have checked a larger range, but my computer crashed.

I also thought the solving the one-dimensional case completely might help, and can see that x\mapsto ax^3 + bx^2 + cx + d is injective if and only if it cannot be written as a(x-A)(x-B)\left(x-\frac{C}a\right)+k for integer a,A,B,C,k – but this isn’t super helpful, far as I can tell. However, the statement f(x,y) is injective is equivalent to asserting that t\mapsto f(m_1t + b_1,m_2t+b_2) is injective for all m_1,b_1,m_2,b_2\in\mathbb Z with not both m equalling 0 – so this could be used to eliminate some cases, if nothing else.


Disclaimer: This is merely a too lengthy comment to fit in the comment section.

I still have no idea about the general degree 3 case, but here is another proof that no polynomial of degree 2 will work:

Write the polynomial P(x,y) of degree 2 as

P(x,y)=\sum_{i+j\leq 2}c_{ij} x^i y^j,\quad c_{ij}\in\mathbb Z

Consider a point (a,b)\in\mathbb Z^2\setminus\{(0,0)\} and the expression


If the coefficient (c_{10}a+c_{01}b) is zero, then t\mapsto P(ta,tb) is an even function. Then


for all t\in\mathbb Z. If c_{10}=c_{01}=0 we can choose any (a,b)\in\mathbb Z^2\setminus\{(0,0)\}. Otherwise (a,b)=(-c_{01},c_{10})\neq(0,0) works. In any case, we have found an infinitude of pairs (ta,tb),(-ta,-tb) contradicting P being injective.

Source : Link , Question Author : Milo Brandt , Answer Author : String

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