A notion of independence for beatty sequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:03:20Z http://mathoverflow.net/feeds/question/23163 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23163/a-notion-of-independence-for-beatty-sequences A notion of independence for beatty sequences SJR 2010-05-01T09:00:55Z 2010-05-02T04:54:45Z <p>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 $x$?</p> <p>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.)</p> <p>Can anyone do $n=3$??</p>