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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, RouchÃ©'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.
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, RouchÃ©'s theorem shows that the number of solutions 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.