Let $K$ be a number field with ring of integers $O_K$. Is there a section of $\mathbf{P}^1_{O_K}$ over $O_K$ whose image is disjoint from $0$, $1$ and $\infty$? If $K=\mathbf{Q}$ this is not possible because any integer $n>1$ is divisible by a prime number. What if $K \neq \mathbf{Q}$?

Such sections are tantamount to solutions of the unit equation $u + u' = 1$ in $O_K^*$. This is indeed impossible for $K = {\bf Q}$, when $O_K = {\bf Z}$ and the only units are $\pm 1$; but there can be such solutions for other number fields $K$, though it is known that in each $K$ there are only finitely many solutions. A "section of ${\bf P}^1_{O_K}$ over $O_K^{\phantom}$" is a $K$point of the projective line, i.e. either $\infty$ or a field element. The "sections" $u$ disjoint from $\infty$ are precisely the algebraic integers, because $u$ "intersects $\infty$ at the prime $\wp$" iff $u$ has negative valuation at $\wp$, and the algebraic integers are precisely the field elements none of whose valuations are negative. Likewise $u$ is disjoint from $0$ iff $u$ has no positive valuation at any $\wp$, and disjoint from $1$ iff $u1$ has no positive valuations. Therefore, $u$ is disjoint from $0$, $1$, and $\infty$ iff both $u$ and $u1$ are units, which is to say iff $(u,1u)$ is a solution of the unit equation. One easy way to get such $(K,u)$ is to make $u(1u)$ a unit in some number field $F$, say $\epsilon$, because then $u$ and $1u$ are themselves algebraic integers in a number field containing $F$ with degree at most $2$ (they're the roots of the monic quadratic polynomial $x^2x+\epsilon$ over $O_F$), and thus units because they divide the unit $\epsilon$. For example, taking $F={\bf Q}$ and $\epsilon = 1$ or $1$ we recover the simplest solutions of the unit equation: the sixth roots of unity $(1 \pm \sqrt{3})/2$, and the golden ratio $(1 \pm \sqrt{5})/2$. More generally, if $x^{m+n}  x^m = \epsilon$ for some unit $\epsilon$ and positive integers $m,n$ then both $x$ and $1x^n$ are units, so we may take $u = x^n$. 

