Pythagorean Equation

Introduction

We're all familiar with the equation 

$x^2+y^2=z^2 (1)$ 

describing the relationship between the sides of a right triangle. In this article, we're particularly interested in integral solutions $(x,y,z)$.

Algebraic Approach

Remark 1: It's sufficient to solve $(1)$ with $gcd(x,y)=1$ 

First, note that we only need to concern ourselves with pairs of $(x,y)$ such that $gcd(x,y)=1$. If $gcd(x,y)=d$, easily $d|z$ and we divide both sides of $(1)$ by $d^2$ to get $x'^2+y'^2=z'^2$ (where x=x'd and similarly for $y,z$). Note that remark 1 also forces $gcd(x,z)=gcd(x,y)=1$. 

Remark 2: It's sufficient to solve $(1)$ with $x \equiv 1 \pmod{2}$ and $y \equiv 0 \pmod{2}$  

Due to remark 1, $x$ and $y$ cannot be both even lest they share the common factor $2$. We further show that they cannot both be odd either. On the contrary, assume that $x$ and $y$ are both odd. So 

$x^2 + y^2 \equiv 2 \pmod{4}$

 leading to $z^2 \equiv 2 \pmod{4}$. This is impossible, as the remainder set mod $4$ is $\{ 0,1,2,3\}$, and the squared set equals $\{0,1,4,9\} \equiv \{0,1,3\}$ and does not contain $2$. 

General solution

Due to remark 2, we assume that $x$ is odd and $y$ is even. From $(1)$,

 $y^2=z^2-x^2=(z-x)(z+x) (2)$

Let $gcd(x-z,x+z)=d$. We have $d|z+x$ and $d|z-x$. Therefore, $d|z+x+(z-x)$ or $d|2z$. Similarly, $d|z+x-(z-x)$ or $d|2z$. Since we assume $gcd(x,z)=1$, $d|2 \rightarrow d=2$. Dividing both sides of $(2)$ by $4$, we get

$\displaystyle (\frac{y}{2})^2 = (\frac{z-x}{2})(\frac{z+x}{2})$

Now, $gcd(\frac{z-x}{2},\frac{z+x}{2})=1$. Easily, if $kt=c^2$ and $gcd(k,t)=1$, $k$ and $t$ must each be squares themselves. So 

$\frac{z-x}{2}=a^2$ and  $\frac{z+x}{2}=b^2$.

So,

                            $a^2+b^2=z$; $b^2-a^2=x$; $2ab=y$. 

Geometrical approach

From

                              $x^2+y^2=z^2$

we reduce to,

$\displaystyle \Big(\frac{x}{z}\Big)^2 + \Big(\frac{y}{z}\Big)^2=1$

So to solve Pythagorean equation, we just need to find a point $(\frac{x}{z},\frac{y}{z})$ with rational coordinates in the unit circle. 

Claim: a rationally sloped line through a rational point on the unit circle also goes through another rational point on the circle.

Proof
Before proving this claim, we will introduce a trigonometric identity

$tan(\theta) = \displaystyle \sqrt{\frac{1-cos(2\theta)}{1+cos(2\theta)}}$ 

We have,

$\displaystyle tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$\displaystyle \frac{2sin(\theta)cos(\theta)}{2cos^2(\theta)} = \frac{sin(2\theta)}{cos(2\theta)+1}$

$\displaystyle \sqrt{\frac{1-cos^2(2\theta)}{(cos(2\theta)+1)^2}} =  \sqrt{\frac{1-cos(2\theta)}{1+cos(2\theta)}}$

We now return to the Pythagorean Equation. Through the initial rational point $A=(-1,0)$, we draw a rationally sloped line cutting the unit circle at another point $B$. We will prove that $B$ is also a rational point. Because the triangle $AOB$ is isosceles, $\angle BCD = 2 \angle BAC$. But because the line $AB$ is rationally sloped, $tan(2\angle BAC) \in \mathbb{Q}$. This leads to

        $\displaystyle \sqrt{\frac{1-cos(2\angle BCD)}{1+cos(2\angle BCD)}} \in \mathbb{Q}$ 

Easily, $\angle BCD \in \mathbb{Q}$. Because $BC=1 \in \mathbb{Q}$, $BD \in \mathbb{Q}$. Easily, $BC \in \mathbb{Q}$. Thus, the $x$ and $y$ coordinates of $B$ are both rational. This proves our claim. 

Lastly, solving two equations

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

simultaneously, we get the same solutions as in the algebraic approach.