Let $A$ be a set of integers with irrational natural density. That is, suppose that


$\lim\_{n\to\infty}\frac{\#(A\cap [-n,n))}{2n}$


exists and is irrational. Denote this value by $\alpha$. Now let $p$ be an odd polynomial with integer coefficients, and define the set $S\_{p}(k) = \{p(i) : i \in [-k, k)\cap\mathbb{Z}\}$. Is is the case that any such $p$ gives


$\lim\_{k\to\infty}\frac{\#(A\cap S\_{p}(k))}{2k} = \alpha$?


A note on the necessity of these conditions: If $\alpha$ is rational, this fails. Take $\alpha = 1/2$, $A = 2\mathbb{Z}$, and $p(x) = 2x$. If $p$ is even, I suspect (but cannot prove) that this also fails. To see this, fix $\alpha$ irrational and construct $A$ in such a way that slightly more of the "mass" of $\alpha$ falls on $A\cap[0,\infty)$ than on $A\cap(-\infty,0)$. Then take $p(x) = x^2$.


I suspect that ergodic theory may be helpful here, but I have not found a good way to think about it yet! Any ideas are welcome.



Irrational natural density set, intersected with odd polynomial
A slight modification of your example shows the answer is no. Let $A$ consist of the even numbers together with a set of odd numbers with irrational density, and take $p(x) = 2 x$.


