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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.

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    $\begingroup$ I think Bellabas should be Belabas. $\endgroup$ Jan 2, 2010 at 4:56

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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.

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This is an old question, but maybe it's worth mentioning that a conjecture of Erdös, Stewart, and Tijdeman [1] says that if $P_s=\{2,3,\ldots,p_s\}$ is the set of the first $s$ primes, then the number of solutions to the $S$-unit equation in $\mathbb Q$ should satisfy $$ \exp(s^{2/3-\epsilon}) \le \text{\# of sols using $P_s$} \le \exp(s^{2/3+\epsilon}) \quad \text{for all $s>s_0(\epsilon)$}. $$ If this is true, and if (as Bjorn suggests) the smaller primes are the ones yielding the most solutions, this would show that if one takes the square root of the number of primes, then the number of solutions decreases by a large (and reasonably calculable) amount.

In the same article, they prove that there are sets $S$ containing $s$ primes so that the number of solutions is roughly $$ \exp\left( \sqrt{\frac{s}{\log s}} \right). $$ Their clever construction take a lot of initial primes and throws in some carefully selected larger primes.

[1] Erdös, P.; Stewart, C. L.; Tijdeman, R. Some Diophantine equations with many solutions. Compositio Math. 66 (1988), no. 1, 37–56. MR0937987

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  • $\begingroup$ You may be interested to know that the result of Erdos, Stewart and Tijdeman was improved in work of Konyagin and Soundararajan (arxiv.org/abs/math/0604453; published paper in JNT), where the $\sqrt{s}$ is improved to $s^{2-\sqrt{2}-\epsilon}$. $\endgroup$
    – Lucia
    Jun 12, 2020 at 1:54
  • $\begingroup$ @Lucia Thanks. That's good to know. Here's the JNT referece: Konyagin, S.; Soundararajan, K. Two S-unit equations with many solutions. J. Number Theory 124 (2007), no. 1, 193–199. MR2321000 $\endgroup$ Jun 12, 2020 at 2:32

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