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

So, the pair of natural numbers is a solution to the equation

.

Everyone knows what the equation looks like: It is a circle C of radius 1 with centre at . 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, and . Suppose that we take any rational number m and look at the line L going through the point and having slope m, (Draw this line!) The line L is given by the equation

L: (point-slope formula)

It is clear from the picture you have drawn that the intersection consists of exactly two points, and one of those points is . We want to find the other one:

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

and 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 must be a solution, since the point is on both C and L. This means that we can divide the polynomial by to find the other root:

So, on dividing by , the quotient polynomial is .

So the other root is the solution of the quotient polynomial

, which means that

. Then, we substitute this value of x into the equation of the line L to find the y-coordinate:

Thus, for every rational number m we get a solution in rational numbers

to the equation

.

On the other hand, if we have a solution in rational numbers, then the slope of the line through and will be a rational number. So, by taking all possible values for m, the process we have described will yield every solution to (except for

which corresponds to the vertical line having slope .

**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 , then our formula becomes 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**

and .

More later…

Nalin Pithwa

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## One Comment

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.