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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.

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?

I also posted this question on stackexchange yesterday, but didn't get an answer.

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I added some more appropriate tags. This has nothing to do with "complex geometry" or "complex analysis" in the sense they're normally understood, and it's recommended to include at least one Arxiv subject class tag. Link to math.SE post: – David Loeffler Oct 23 '11 at 9:17
up vote 7 down vote accepted

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.

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This is very elegant. Some questions. 1. Why do you require $y$ not to lie in $\mathbf{R}$? 2. To apply Rouche's theorem, one has to verify that $\vert x(1-x)(x^n+ux+v)-1\vert + \vert x(1-x) ( x^n+u^\prime x +v^\prime) -1 \vert \geq \vert x(1-x)(\langle u\rangle x +\langle v\rangle)\vert$, where $x$ lies in a suitable neighborhood of $y$, and $n$ is big enough and I use some non-standard notation for the nearest integer and fractional part, right? (That's a question just to be sure I understand the way you apply the theorem.) – Bana Oct 23 '11 at 11:03
1. I was thinking of $1,y$ as an $\mathbf R$-basis, but $y\ne 0,1$ is indeed sufficient (strictly speaking, I need to exclude $(1\pm i\sqrt3)/2$ as well to get $|y|>1$). 2. Actually, you need to consider a contour around $y$. There is an asymmetric form of Rouché's theorem which lets you use the fact that the derivative is large. – user2035 Oct 23 '11 at 14:09

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