## 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

$(a/c)^{2}+(b/c)^{2}=1$

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

$x^{2}+y^{2}=1$.

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

$x^{2}+(m(x+1))^{2}=1$

$x^{2}+m^{2}(x^{2}+2x+1)=1$

$(m^{2}+1)x^{2}+2m^{2}x+(m^{2}-1)=0$

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

$x^{2}+y^{2}=1$.

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

$(a,b,c)=(u^{2}-v^{2},2uv,u^{2}+v^{2})$.

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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