Quadratic Curves and Surfaces

Our first problem is that of finding Pythagorean triples, that is integers a,b,c, such that a^2 + b^2 = c^2, for example a=3, b=4, c=5. By dividing both sides by c^2 and letting x=a/c and y=b/c, we obtain the equation

x^2 + y^2 = 1.

This is the equation of a circle of radius 1 centered at the origin and we are trying to find points on it that have rational coordinates.

Now this equation has a "trivial" solution, namely x=0 and y=1. We can use this trivial solution to find all other rational solutions as follows. Consider the line of slope m passing through this point, namely

y = mx + 1


We want to find where this line intersects the circle. Substituting we have

x^2 + (mx + 1)^2 = 1
Now this is a quadratic equation in x and in general we would not expect to find rational solutions, but in this case we already know that x=0 is a root and hence the other root must also be rational. Expanding we obtain

(1 + m^2)x^2 + 2mx = 0 so
x = 0 or x = -2m/(1+m^2) [and hence y = (1-m^2)/(1+m^2)].

To recover our Pythagorean triples from these rational solutions, we take c to be the least common denominator of x and y, and let a=cx and b=cy. For example, if m = -1/4, then x = 8/17 and y = 15/17, and a=8, b=15, and c=17. This method will construct any Pythagorean triple.

This technique may be applied to any quadratic equation in any number of variables. If one can find one rational solution, there must be infinitely many and they can all be found by this technique.

Exercise: Find a parametric representation for the rational solutions of

x^2 + y^2 + z^2 = 1