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I've read an interesting article, math.NT/0409456 where you're just trying to solve a simple problem:

For a given (finite) set of primes S find all solutions to an equation a + b = c with the condition that all prime divisiors of integers a, b, c must be in S.

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 R --> Spec ZZ \ S. The article then proves that the number of solutions to the equation is finite by proving that the number of these sections is finite.

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?

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  • $\begingroup$ Downvoted for vagueness: I'm not sure at all what ilya n. is asking for here. $\endgroup$
    – JSE
    Commented Oct 24, 2009 at 5:19
  • $\begingroup$ Hope it's better now. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 24, 2009 at 6:20

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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 the outstanding problem in Diophantine analysis.

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  • $\begingroup$ Wow, this is an amazing connection to a hard problem. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 31, 2009 at 10:40
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The original problem you mention is also known as the "S-unit equation": see Wikipedia's page (http://en.wikipedia.org/wiki/S-unit_equation#S-unit_equation). 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).

I haven't yet figured out what's going on with the connection to schemes, but it's pretty nifty that there is one!

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  • $\begingroup$ I can't resist explaining the connection to schemes. So, first let's consider the scheme Spec ZZ (if you don't know what this is, then thin about it as just the collection of primes numbers). Now what's an integer? Simple -- it's a function on this thing. Ok, what's a rational? Also easy -- it's a rational function, that is something with poles. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 23, 2009 at 22:46
  • $\begingroup$ Now I want to solve an equation x + y = 1 where x and y have no primes other than the ones in S. In other words, they have only zeroes or poles on S when you consider them as rational functions. Ok, so let's say I have an x with this property. Then y is uniquely defined and I need it to have no poles or zeroes outside of S. This means x must have no poles or ones outside of S as well. So, our problem is reformulated as finding x such that at every prime p the value of x is neither 0 nor 1 nor infinity. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 23, 2009 at 22:49
  • $\begingroup$ In other words, x should be an element of projective space minus three points over every prime p for primes not in S. This can be combined together to say that x is a section of a map (P^1-(0, 1, infty)) over (Spec ZZ - S). $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 23, 2009 at 22:50
  • $\begingroup$ I was wrong about counting, btw. It's about proving things mostly. Your answer is very relevant. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 23, 2009 at 22:52
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This sounds to me like it is related to some recent work of Lagarias and Soundararajan:

http://arxiv.org/pdf/0911.4147

Their paper has some relations to the ABC conjecture and to GRH. Hope this helps.

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