# How to find integer solutions of an equation using approximation methods?

If I have a function called $f(x)$ that have several roots, integers and not integers. How can I find just the integer ones by approximations methods?

A simple example would be $\sin(\frac{2x\pi}{3})$ that have solutions like $0, 1.5, 3, 4.5,\ldots$

So, how could I find the integer solutions using approximations methods like Newton-Raphson? Is there a way to do this? Which method or methods can I use?

PS-1: I would like study more complicated functions with no trivial solutions, but this simple example can explain what I'm looking for.

PS-2: I'm not looking for a computational algorithm but a theorem that shows the iterable steps will converge to integer solutions. I'd find something similar at the division algoritm.

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The approximation methods don't care for integers/non-integers at all. Unless you are talking about some special method for specific problems, this makes no sense. –  vonbrand Apr 16 '13 at 23:24
Maybe it isn't. Probably the approximation steps will walk over any numbers, but the final result should just return integer solutions. I think we have something like these, I think I have heard about something like that before. But I couldn't find a reference yet. –  GarouDan Apr 16 '13 at 23:29
This may sound stupid, but why don't you just brute-force (if you have the chance). Another option is, since you have already mentioned "complicated functions," you may want to try a genetic algorithm and quickly find what you are looking for. –  Lord Soth Apr 16 '13 at 23:46
@LordSoth, unfortunally genetic, brute force, is not really I'm looking for. I'm needing a math sentence like the Newton-Rapshon that I described above that works. I need this to analize the function that I'm studying analytically. –  GarouDan Apr 16 '13 at 23:50
@Garou: The algorithm "Run Newton's method but pick a new starting point whenever it looks like you're getting close to something not an integer" does converge only to integer solutions! –  Hurkyl Apr 17 '13 at 0:02