# A notion of independence for beatty sequences

Let $\lfloor\cdot\rfloor$ denote the floor function. Is it true that for every positive integer $n$, there are real algebraic numbers $\alpha_1\ldots\alpha_n$ such that for every non-zero polynomial $p\in\mathbb{Z}[x_1\ldots x_n]$, the equation $$p(\lfloor \alpha_1 x\rfloor \ldots \lfloor \alpha_n x\rfloor )=0$$ has only finitely many solutions in integers $x$?

The statement is true for $n=2$. In fact, no equation $p(x,\lfloor 2^{1/3}x \rfloor )=0$ can have infinitely many integer solutions. (Proof-Sketch: Suppose that, on the contrary, the equation had infinitely many solutions. Then $p(x,y)$ would have an irreducible factor $q(x,y)$ with infinitely many integer zeros, and the leading homogeneous part of $q(x,y)$ would be divisible by $x^3-2y^3$. But it is known that any irreducible polynomial $q\in\mathbb{Z}[x,y]$ with infinitely many integer zeros has leading homogeneous part a constant multiple of a power of a linear or quadratic form: See Theorem 21, Chapter 22 of Mordell's Diophantine Equations.)

Can anyone do $n=3$??

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This is similar (but certainly different) to my work with Ron Graham, ["Can you hear the shape of a Beatty sequence?"][1]. [1]: arxiv.org/abs/0809.0004 – Kevin O'Bryant May 1 '10 at 18:45
Thank you! I wasn't aware of your paper or of the open problems you mention in it. Do you know of any references that specifically address the identity problem for generalized polynomials? – SJR May 2 '10 at 4:59