Pythagorean Triples and the Unit Circle — for RMO

In the previous blog, we described all solutions to a^{2}+b^{2}=c^{2} in whole numbers a, b and c. If we divide this equation by c^{2}. we obtain


So, the pair of natural numbers (a/c,b/c) is a solution to the equation


Everyone knows what the equation x^{2}+y^{2}=1 looks like: It is a circle C of radius 1 with centre at (0,0). We are going to use the geometry of the circle C to find all the points on C whose xy-coordinates are rational numbers. Notice that the circle has four obvious points with rational coordinates, (\pm 1, 0) and (0,\pm 1. Suppose that we take any rational number m and look at the line L going through the point (-1,0) and having slope m, (Draw this line!) The line L is given by the equation
L: y=m(x+1) (point-slope  formula)

It is clear from the picture you have drawn that the intersection C \bigcap L consists of exactly two points, and one of those points is (-1,0). We want to find the other one:

To find the intersection of C and L, we need to solve the equations:

x^{2}+y^{2}=1 and y=m(x+1) for x and y. Substituting the second equation into the first and simplifying, we need to solve




This is just a quadratic equation, so we could use the quadratic formula to solve for x. But, there is a much easier way to find the solution. We know that x=-1 must be a solution, since the point (-1,0) is on both C and L. This means that we can divide the polynomial by x+1 to find the other root:

So, on dividing (m^{2}+1)x^{2}+2m^{2}x+(m^{2}-1) by x+1, the quotient polynomial is (m^{2}+1)x+(m^{2}-1).

So the other root is the solution of the quotient polynomial

(m^{2}+1)x+(m^{2}-1)=0, which means that

x=(1-m^{2})/(1+m^{2}). Then, we substitute this value of x into the equation y=m(x+1) of the line L to find the y-coordinate:

y=m(x+1)=m( ((1-m^{2})/(1+m^{2})+1) ) =(2m)/(1+m^{2}) Thus, for every rational number m we get a solution in rational numbers

((1-m^{2})/(1+m^{2}),(2m)/(1+m^{2})) to the equation


On the other hand, if we have a solution (x_{1},y_{1}) in rational numbers, then the slope of the line through (x_{1},y_{1}) and (-1,0) will be a rational number. So, by taking all possible values for m, the process we have described will yield every solution to x^{2}+y^{2}=1 (except for

(-1,0) which corresponds to the vertical line having slope m=\infty).

How is this formula for rational points on a circle related to our formula for Pythagorean triples (in the previous blog)?

If we write the rational number m as a fraction v/u, then our formula becomes (x,y)=((u^{2}-v^{2})/(u^{2}+v^{2}), (2uv)/(u^{2}+v^{2})) and clearing the denominators gives the Pythagorean triple


This is another way of describing Pythagorean triples, although to describe only the primitive ones would require some restrictions on u and v. You can relate this description to  the formula in the earlier blog by setting

u=(s+t)/2 and v=(s-t)/2.

More later…

Nalin Pithwa



One Comment

  1. Posted October 26, 2014 at 6:09 am | Permalink | Reply

    Just after the first step we needed rational points on the unit circle. We may write the general solution as x=1cosΔ y=1sinΔ where both sine and cosine are rational. Converting it to tan(Δ/2)=m we get x=(1-m^2)/(1+m^2) and y=2m/(1+m^2) where m is any positive rational number.

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