Do the solutions to the unit equation lie dense in the complex numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:02:21Z http://mathoverflow.net/feeds/question/78876 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78876/do-the-solutions-to-the-unit-equation-lie-dense-in-the-complex-numbers Do the solutions to the unit equation lie dense in the complex numbers Bana 2011-10-23T08:25:28Z 2011-10-23T10:30:08Z <p>Let $S\subset \overline{\mathbf{Q}}\subset \mathbf{C}$ be the set of solutions to the unit equation, i.e., $S$ consists of algebraic integers $a$ such that $a$ and $1-a$ are units in the ring of algebraic integers.</p> <p>Let $U$ be a non-empty open subset in the Euclidean topology on $\mathbf{C}$. </p> <p>Does $U$ contain infinitely many solutions to the unit equation. That is, does the intersection $S\cap U$ contain infinitely many elements?</p> <p>I also posted this question on stackexchange yesterday, but didn't get an answer.</p> http://mathoverflow.net/questions/78876/do-the-solutions-to-the-unit-equation-lie-dense-in-the-complex-numbers/78880#78880 Answer by a-fortiori for Do the solutions to the unit equation lie dense in the complex numbers a-fortiori 2011-10-23T09:45:20Z 2011-10-23T10:30:08Z <p>If $f\in\mathbf Z[X]$ is any monic polynomial, the solutions of $x(1-x)\cdot f(x)=1$ are solutions of the unit equation. Take some $y\in U\setminus\mathbf R$. Since the substitution $z\mapsto1/(1-z)$ leaves $S$ invariant, we may assume $|y|>1$. For $n$ given, choose $u,v\in\mathbf R$ such that $y(1-y)\cdot(y^n+uy+v)=1$. Now if $n$ is sufficiently large, Rouché's theorem shows that the number of solutions in a suitable neighbourhood of $y$ in $U$ does not change if we replace $u$ and $v$ by the nearest integers. Hence, $S\cap U$ is nonempty. Since $U$ was arbitrary, this implies that $S\cap U$ is infinite.</p>