MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Downvoted for vagueness: I'm not sure at all what ilya n. is asking for here. – JSE Oct 24 '09 at 5:19
Hope it's better now. – Ilya Nikokoshev Oct 24 '09 at 6:20
up vote 1 down vote accepted

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.

share|cite|improve this answer
Wow, this is an amazing connection to a hard problem. – Ilya Nikokoshev Oct 31 '09 at 10:40

The original problem you mention is also known as the "S-unit equation": see Wikipedia's page ( 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!

share|cite|improve this answer
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. – Ilya Nikokoshev Oct 23 '09 at 22:46
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. – Ilya Nikokoshev Oct 23 '09 at 22:49
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). – Ilya Nikokoshev Oct 23 '09 at 22:50
I was wrong about counting, btw. It's about proving things mostly. Your answer is very relevant. – Ilya Nikokoshev Oct 23 '09 at 22:52

This sounds to me like it is related to some recent work of Lagarias and Soundararajan:

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

share|cite|improve this answer
Thanks, very interesting paper! – Ilya Nikokoshev Nov 24 '09 at 19:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.