# Finding integer points

1. Nov 24, 2006

let be a 2 dimensional symmetryc form:

$$z=f(x,y)=ax^{2}+bxy+cy^{2}$$

depending on the values of a,b and c we'll have an elipse , parabole and hyperbola or circumference,my question is are there any geommetrical methods to find integer points (x,y) satisfying the equation z=constant ??

2. Nov 24, 2006

### matt grime

Rational points on curves. A well known area of research, just f**king well google it.

3. Nov 24, 2006

### mathwonk

read serre's book "course on arithmetic".

4. Nov 25, 2006

### Playdo

Some of the fun of math as a hobby is not having to read the book. The question is interesting K-bad. Consider this function.

(2) $$g(x,y,\beta) = 1-tan^{\alpha}(\frac{\pi}{2}(1 - \frac{1}{2}(sin^{2\beta}(\pi x)+sin^{2\beta}(\pi y))))$$

This function only equals zero when both x and y are integers. Imagine your curve as the trajectory of a particle moving through the plane. The arc length of the curve in the first quadrant for instance has some value. Now consider the same trajectory taken over the "dimpled plane" g(x,y). If the trajectory comes "close to" an integer point then

(3) $$\int{f(x,y)g(x,y,\beta)}ds > \int{f(x,y)}ds$$

So, by adjusting beta you can get a very strong hint that your curve may strike integer points. The larger you set beta the tighter the hole around the integer points gets. So theoretically you could tighten it down so far that only curves that have integer points will show a difference in Equation 3. Try it! Mess around with beta and your favorite plane curve. Try the class of curves y = N/x.

Now the more technical side of this is that you are going to have to use numerical methods and that could be kicker. I would solve (3) by converting everything to infinite series (convergence is now an issue) and then integrating term by term. For the factorization method on the class y = N/x the limits of integration are x = 1 to x = root N.

Clearly the technology here if you will is understanding the functions that can produce a dimpled plane, the arc length and hyperbolas, not too advanced. But you can use modular sieves to create very sparse "integer points". The standard intger lattice is x = u, y = v. But some of the modular dimpled planes are like x = au+ r, y = bv + s. You get this information almost for nothing from Div(N,a) and Mod(N,a).

I'm just beginning a thorough numerical study of this idea to see what the distributions look like. It would seem that again theoretically I would make the probability of being wrong about a number being prime or composite as small as computing power would allow. Also if you suspect that a curve N = xy has an intger point using this method, you can divide the domain into two parts, run the same test, then four etc. So the search for the actual factor becomes probabilistic and is probably quite fast in general.

This method requires only these undergraduate math components to begin to investigate: hyperbolas, trig functions, arc length, modular arithmetic, infinite series, and basic algorithms.

Have fun man. Don't topple the RSA system on a whim though :!!)

Last edited: Nov 26, 2006
5. Nov 25, 2006

### d_leet

No it doesn't. It equals one when both x and y are integers unless you mean for both of those sines to be cosines.

6. Nov 26, 2006

### robert Ihnot

Karlisbad: is are there any geommetrical methods to find integer points (x,y) satisfying the equation z=constant ??

Good Heavens! The conics were known for many centuries before Analytical Geometry. Menaechmus, 375-325 BC is considered the first person to define the forms. Euclid worked on the circle and Archimedes using "triangle integration" found the area of the parabola.

7. Nov 26, 2006

### Playdo

Yes, thank you of course. Wrong form. I will repost that.

8. Nov 26, 2006

### Playdo

Yes, thank you of course. Wrong form. I will repost that. It is done.

9. Nov 27, 2006

### Playdo

deleted until further notice

10. Dec 14, 2006

if we have $$Nx^{2}-y^{2}=1$$ then if N is a perfect square the solution is trivial..if N isn't a perfect square perhaps we could "split" te Pell equation in the form:
$$(Ax-y)(Ax+y)+\alpha x^{2}=1$$ where $$0<\alpha <1$$
$$x=\sum_{n=0}^{\infty}c_{n}\alpha ^{n}$$ $$y=\sum_{n=0}^{\infty}d_{n}\alpha ^{n}$$