Let $K$ be a number field, and let $S\_x$ denote the set of primes of norm at most $x$. Is it possible to find a smaller set of places $T\_x\subset S\_x$ so that a lot of the solutions of the $S\_x$-unit equation $a+b=1$ for $a,b\in S\_x$ are solutions of the $T\_x$-unit equation?


Here's a possible precise statement (although I'd be interested in other formulations as well): Does there exist a constant $0<c<1$, depending on $K$ (but not $x$), so that for each $x$, there is a $T\_x\subset S\_x$ with $|T\_x|\le\sqrt{|S\_x|}$ so that the number of solutions to the $T\_x$-unit equation is at least $c$ times the number of solutions of the $S\_x$-unit equation?


I'm interested in this mostly by analogy: Belabas and Gangl have a bound for the set of places of a number field one must check in order to compute $K\_2$ of the ring of integers. It would be interesting to know if one could at least get a pretty good approximation for $K\_2$ by looking at a much smaller set of places.



S-unit equation and small sets of places
I think that the precise statement you ask for is false, but I have a feeling that current bounds on number of solutions to $S$-unit equations are not good enough to prove this, even for $K=\mathbf{Q}$. 


What I can prove is that it is false if you require $T\_x$ to be the subset of the specified size consisting of the *smallest* primes of $S\_x$. (And it seems unlikely that using larger primes would be much better; this is why I think that the precise statement is false.)


**Proof:** Let me assume $K=\mathbf{Q}$ for simplicity. Let $f(x)$ be the number of solutions to the $S\_x$-unit equation. We have $T\_x=S\_y$ where $y=x^{1/2+o(1)}$, so if $c$ exists, we would have $f(x^{1/2+o(1)}) \ge c f(x)$ for all $x$. Iterating this shows that $f(x)$ is bounded by a polynomial in $\log x$ as $x \to \infty$. But $f(x) \ge x-1$ because of the solutions $a+(1-a)=1$ for $a=2,\ldots,x$. This is a contradiction.


