# Do the solutions to the unit equation lie dense in the complex numbers

Let $S\subset \overline{\mathbf{Q}}$ 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.

Let $U$ be a non-empty open subset in the Euclidean topology on $\mathbf{C}$.

Does $U$ contain infinitely many solutions to the unit equation. That is, does the intersection $S\cap U$ contain infinitely many elements?

Since there werent't any replies, I also asked this question on Mathoverflow.

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mathoverflow.net/questions/78876/… to be precise, where there is now an answer posted. – Gerry Myerson Oct 23 '11 at 11:49

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.