Solving "a, b, a+b have given divisors" problem - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T18:24:06Zhttp://mathoverflow.net/feeds/question/2142http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/2142/solving-a-b-ab-have-given-divisors-problemSolving "a, b, a+b have given divisors" problemIlya Nikokoshev2009-10-23T18:14:06Z2009-11-24T02:15:41Z
<p>I've read an interesting article, <a href="http://arxiv.org/abs/math.NT/0409456" rel="nofollow">math.NT/0409456</a> where you're just trying to solve a simple problem:</p>
<blockquote>
<p>For a given (finite) set of primes S find all solutions to an equation <code>a + b = c</code> with the condition that all prime divisiors of integers a, b, c must be in S.</p>
</blockquote>
<p>and this problem turns out to be very geometric. It turns out (and I tell that in comments below) you're actually dealing with sections of certain projective morphism of schemes <code>R --> Spec ZZ \ S</code>. The article then proves that the number of solutions to the equation is finite by proving that the number of these sections is finite.</p>
<p>Is there kind of general theory or other methods to prove things about sections of these maps? What is the intuition used here? Would there be a way to count these solutions?</p>
http://mathoverflow.net/questions/2142/solving-a-b-ab-have-given-divisors-problem/2205#2205Answer by Alison Miller for Solving "a, b, a+b have given divisors" problemAlison Miller2009-10-23T22:31:55Z2009-10-23T22:31:55Z<p>The original problem you mention is also known as the "S-unit equation": see Wikipedia's page (<a href="http://en.wikipedia.org/wiki/S-unit_equation#S-unit_equation" rel="nofollow">http://en.wikipedia.org/wiki/S-unit_equation#S-unit_equation</a>). There's a fair amount of literature approaching this problem from a more classical direction, but only about proving that there are finitely many solutions/bounding the number of solutions, not counting the solutions (which seems to be pretty hard).</p>
<p>I haven't yet figured out what's going on with the connection to schemes, but it's pretty nifty that there is one! </p>
http://mathoverflow.net/questions/2142/solving-a-b-ab-have-given-divisors-problem/2878#2878Answer by engelbrekt for Solving "a, b, a+b have given divisors" problemengelbrekt2009-10-27T19:59:23Z2009-10-27T19:59:23Z<p>This is bound to be pretty hard! Knowing the prime divisors of a,b,c implies knowing the radical rad(abc) (the product of the distinct primes dividing abc), and then knowing the solutions a,b,c of a + b = c in positive integers, you would know the largest c with a,b,c coprime (pairwise or not, doesn't matter). So you would be in a position to settle the ABC conjecture c \leq C(epsilon)rad(abc)^{1 + \epsilon} if you could handle all those S-unit equations. The ABC conjecture is <em>the</em> outstanding problem in Diophantine analysis.</p>
http://mathoverflow.net/questions/2142/solving-a-b-ab-have-given-divisors-problem/6645#6645Answer by Ben Weiss for Solving "a, b, a+b have given divisors" problemBen Weiss2009-11-24T02:15:41Z2009-11-24T02:15:41Z<p>This sounds to me like it is related to some recent work of Lagarias and Soundararajan:</p>
<p><a href="http://arxiv.org/pdf/0911.4147" rel="nofollow">http://arxiv.org/pdf/0911.4147</a></p>
<p>Their paper has some relations to the ABC conjecture and to GRH. Hope this helps.</p>