# Chapter three: pythagorean array and unit circle

Time：2020-12-9

We describeda^2 + b^2 = c^2All integer solutions ofa，b，c, ifc^2Divide by this equation

(\frac{a}{c})^2 + (\frac{b}{c})^2 = 1

So rational numbers are right(\frac{a}{c},\frac{b}{c})It’s an equationx^2 + y^2 = 1The solution of

equationx^2 + y^2 = 1Represents a circle C with a center at (0, 0) radius of 1.

### How to find the points on circle C where x and Y coordinates are rational numbers from the perspective of set

There are four distinct points on the circle with rational coordinate: (1,0), (- 1,0), (0,1), (0, – 1)

Suppose we take any (rational) number m and observe the straight line L with the slope of (- 1,0) point m. The equation of straight line L is as follows:y = m(x + 1)The intersection of C and l consists of two points, one of which is (- 1,0). Let’s find the other.

In order to find the intersection of C and l, we need to solve the equations about X and y

x^2 + y^2 = 1\\
y = m(x +1)

The second equation is substituted into the first equation and simplified

(m^2 + 1)x^2 + 2m^2x + (m^2 – 1) = 0

It can be obtained by the root formula of quadratic equationx。 However, there is an easier way to solve the equation. Since (- 1,0) is on C and l, sox = -1It must be a solution. Therefore, it is availablex + 1Divide by a quadratic polynomial to find the other root. So, the other root is the equation(m^2 + 1)x + (m^2-1) = 0This means that

x = \frac{1 – m^2}{1 + m^2}

The equation that takes the value of X to the line Ly=m(x+1)Come and askyCoordinates:

y = m(x +1) = m(\frac{1-m^2}{1+m^2}+1) = \frac{2m}{1 + m^2}

So for every rational numbermThe equation is obtainedx^2 + y^2 = 1A rational number solution of(\frac{1-m^2}{1+m^2},\frac{2m}{1 + m^2})(i.e. the formula of all rational number solutions on a circle)

How does the rational number formula on a circle relate to Pythagorean arrays? If there will be a rational numbermWrite in fractions\frac{v}{u}Then the formula becomes

(x,y) = (\frac{u^2-v^2}{u^2-v^2},\frac{2uv}{u^2 + v^2})

Eliminating the denominator gives a Pythagorean array

(a,b,c) = (u^2-v^2,2uv,u^2+v^2)