Linear polynomials in units of number fields I would be thankful for a reference to any result that says "how often" an equation of the form $$c_1x_1 + c_2x_2 + ... + c_nx_n = 0,$$
where $n$ is fixed, $c_1, ..., c_n \in \mathcal{O}_K$ are arbitrarily given algebraic integers of a fixed number field $K$ and $x_1, ..., x_n$ vary in the set of units of $\mathcal{O}_K$, has/does not have a solution. 
Thank you! 
 A: There's a paper of Evertse, Gyory, Stewart, and Tijdeman in which they prove that for "most" (in an appropriate sense) $a,b\in K$, the equation $ax+by=1$ has at most 2 soutions in units $x,y\in\mathcal{O}_K^*$.  The proof is to assume that there are 3 solutions $(x_i,y_i)$, $1\le i\le 3$, use those three equations to eliminate $a$ and $b$, yielding a sum of units that are products of the $x_i$ and $y_i$. Then results of Schlickewei, Evertse, ... (which are based on Schmidt's subspace theorem) imply that there are only finitely many solutions in which no subsum vanishes. It remains to untangle what happens if some subsum does vanish, which isn't so bad for $ax+by=1$, but becomes combinatorially very complicated when there are more variables. In any case, this gives you an idea of how one attacks this sort of problem, where one wants to know that most choices of coefficients give few solutions in units.
Addendum: First, as GH noted, Gyory has 100+ papers, not 1000+ (162 at current MathSciNet count). Second, I tracked down the references:
MR0939471 Evertse, J.-H.; Győry, K.; Stewart, C. L.; Tijdeman, R., On S-unit equations in two unknowns., Invent. Math. 92 (1988), no. 3, 461–477. 
The generalization to $n$ variables is the following: 
MR2093164 Evertse, Jan-Hendrik, Linear equations with unknowns from a multiplicative group whose solutions lie in a small number of subspaces. Indag. Math. (N.S.) 15 (2004), no. 3, 347–355. 
The conclusion of the paper is that for most choices of coefficients, the solutions in units lie in at most $2^n$ proper linear subspaces of $K^n$.
A: As a partial answer, one can show, that, given a number field $K$, there exist infinitely many $c_1, c_2, c_3 \in \mathcal{O}_K$, such that the principal ideals $c_1\mathcal{O}_K, c_2\mathcal{O}_K, c_3\mathcal{O}_K$ are pairwise coprime and the equation 
$$c_1x_1 + c_2x_2 + c_3x_3 = 0$$
has no solutions with $x_j \in \mathcal{O}_K^*$. 
It suffices to show that there exists an ideal $\mathfrak{m} \subset \mathcal{O}_K$ such that the image of $\mathcal{O}_K^*$ in the residue ring $\mathcal{O}_K/\mathfrak{m}$ is a proper subgroup of the multiplicative group of the latter (then one can choose $c_1$ to be any number in the image, $-c_2$ - any number in the complement of the image and $c_3$ to be the rational prime number $p \in \mathfrak{m}$ to conclude that the equation obtained in this way does not have solutions modulo $\mathfrak{m}$).
That one can always find (many) such $\mathfrak{m}$ seems very natural to expect, however the following is a quite indirect argument that probably should be (much?) simpler: the order of the ray class group of a chosen modulus $\mathfrak{m}$ is a product of the class number $h_K$ and the index $[(\mathcal{O}_K/\mathfrak{m})^*: (\mathcal{O}_K^*/\mathfrak{m})]$ (as explained, for instance, here: https://math.stackexchange.com/questions/74206/ray-class-group ). The orders of the ray class groups of $K$, on the other hand, are arbitrarily large, since the Galois group of each abelian extension of $K$ is isomorphic to some ray class group (and there exist abelian extensions of $K$ of arbitrarily large degree).
