In this expository article, we wish to investigate certain Diophantine equations arising from the Pythagorean theorem for integers and look at ways of investigating some of them. We also offer a proof of Fermat’s Last Theorem for the case $n=4$. The contents in this document are neither new nor original; it is merely an attempt to test my own understanding of the material.

The equation of the form $x^2+y^2=z^2$

The Pythagoras theorem tells us a fact about right angled triangles: that if two side lengths of a right angled triangle (other than that of the hypotenuse) be $x$ and $y$ and the hypotenuse be of length $z$, then $\displaystyle x^2+y^2=z^2$. This motivates us to find all positive integral solutions of the form $\displaystyle x^2+y^2=z^2$.

More specifically, we shall investigate equations of the form $\displaystyle x^2+y^2=z^2$ where $\gcd(x,y,z)=1, x,y,z>0$.

Such triples $(x,y,z)$ are known as primitive Pythagorean triples.

Two lemmas are in order.

Lemma 1: In a Pythagorean triple,one and only one of $x$ and $y$ is even.

Proof:Note that $\gcd(x,y,z)=1$ by the definition of a primitive Pythagorean triplet. If both are even, $2|z\implies \gcd(x,y,z)\ge 2$. If both are odd, $\displaystyle x^2+y^2 \equiv 2 \bmod 4$, a contradiction as any square is either $0$ or $1 \bmod 4$ .So one of them is even, the other is odd.

Lemma 2: If $a,b,c\in \mathbb{N}, \gcd(a,b)=1$ and $ab=c^n$, then there exist positive integers $e , f$ such that $e^n=a,f^n=b$.

${\bf Proof}$: The proof is left to the reader.However, looking at the prime representation of $a$ and $b$ is helpful.

An Important Theorem

Theorem 1:The solutions to the equation $x^2+y^2=z^2$ where $x,y,z>0$, $\gcd(x,y,z)=1$, is given by

$\null \boxed{x=2st,y=s^2-t^2,z=s^2+t^2}$

for some naturals $s,t, s>t>0,\gcd(s,t)=1$ and $s \not \equiv t \bmod 2$.

Proof: $x^2=z^2-y^2\implies$  $\displaystyle \frac{x^2}{4}=\frac{z-y}{2}\cdot\frac{z+y}{2}$. Note that $\gcd(\frac{z-y}{2},\frac{z+y}{2})=1$

If not, let $\gcd(\frac{z-y}{2},\frac{z+y}{2})=d$. Then $d|\frac{z-y}{2}+\frac{z+y}{2}=z$ and $d| \frac{z+y}{2}-\frac{z-y}{2}=y$.

Thus $d$ divides $y$ and $z$. As $x^2+y^2=z^2$, $d$ also divides $x$ which forces $\gcd(x,y,z)\ge d$ which is not true as their gcd is $1$ .

Using lemma 2,

$\frac{z+y}{2}=s^2$ and $\frac{z-y}{2}=t^2$ for some positive integers $s,t$.

So, $z=s^2+t^2$ and $y=s^2-t^2$, $z^2-y^2=4s^2t^2\implies x=2st$. Note that $y>0$ means $s^2-t^2>0$ which in turn implies that $s>t>0$.Further, $\gcd(\frac{z-y}{2},\frac{z+y}{2})=1$ forces $\gcd(s,t)$ to be 1 as well.

QED.

Two More Theorems

Theorem 2: There is no solution to $x^4+y^4=z^2$ in positive integers $x,y,z$.

The proof of this theorem will need some work. We will see repeated application of theorem 1 and the two lemmas preceding it.

Proof :We prove this by contradiction.

Suppose that the equation has a solution $(x,y,z)$.

Further, let $\gcd(x,y)=d$. So there exist natural numbers $x_0,y_0$ such that $\gcd(x_0, y_0)=1$ so that $\null x=dx_0$ , $y=dy_0$.

That implies $\null z=d^{2}z_0$ for some $\null z_0$ in naturals.

We now have $\null x_0^4+y_0^4=z_0^2$ such that $\null \gcd(x_0, y_0,z_0)=1, x_0,y_0,z_0>0$.

So we now have a Pythagorean triple $(x_0^2,y_0^2,z_0)$ and so we have (assuming $\null x_0$ is even) $\null \boxed{x^2_0=2st, y_0^2= s^2-t^2,z_0=s^2+t^2}$for some positive naturals $\null s,t, s>t$ with $\null \gcd(s,t)=1$ .

$y_0$ and $z_0$ are odd by lemma 1.

If $\null t$  is odd, $\displaystyle y_0^2\equiv 1\equiv 0-1 \equiv 3\bmod 4$, a contradiction ( note that if t is odd, s is even). So,$t$ is even.

Let $\null t=2k$. Then $\null x_0^2=2s(2k)\implies \exists u,w \in \mathbb{N}$ such that $\null s=u^2, k=w^2$.

Consider the equation $\null y_0^2+t^2=s^2, \gcd(y_0,t,s)=1$(note that $\null \gcd(s,t)=1$ implies that $\gcd(y_0,t,s)$ is 1. So, there exist natural numbers $\null m, n,m>n>0,\gcd(m,n)=1$ such that $\null \boxed{t^2=2mn,y_0=m^2-n^2,s=m^2+n^2,m\not\equiv n \bmod 2}$.

Once again, $\null m=m_1^2$ and $\null n=n_1^2$ for some natural numbers $\null m_1,n_1$ using the previous arguments.

So, $\null s=u^2=(m_1^2) ^2 +(n_1^2)^2$ . Note that this makes $\null (m_1,n_1,u)$ a solution set to our equation.

$\null 0 i.e for each $\null x_0$, there is a smaller $\null m_1$ which is part of the solution of the equation. $\null x_0$ being finite, $\null m_1=0$ at some stage, a contradiction.This is the Fermat’s method of infinite descent.

So the equation $\null \displaystyle x^4+y^4=z^2$ has no solution in natural numbers.

Note that this implies that that there is no solution to the equation $\null x^4+y^4=z^4$, the Fermat’s Last Theorem for the exponent $\null 4$.

Theorem 3: There is no solution to the equation $\null\displaystyle x^4-y^4=z^2$ in natural numbers.

Proof: Imitating theorem 1’s proof, we consider that case when $\null x_0^4-y_0^4=z_0^2$ for coprime $\null x_0,y_0,z_0$.

So now there are two cases. If $\null y_0$ is odd,for some natural numbers $\null m,n$, $\null \boxed{ x_0^2=m^2+n^2,y_0^2=m^2-n^2,z_0=2mn}$ where $\null \gcd(m,n)=1$ and $\null m>n>0$ and $\null m\not\equiv n \bmod 2$.

Multiplying the first two equations, we get $\null m^4-n^4=(x_0y_0)^2$ i.e. $\null \gcd(m,n,x_0y_0)=1$, $\null 0

: the Fermat’s Method of infinite descent shows that there is a contradiction!

We are left with the case $\null y_0$ is even.

Then for some natural numbers $\null m,n$,$\null \boxed{ x_0^2=m^2+n^2,z_0=m^2-n^2,y_0^2=2mn}$ where $\null \gcd(m,n)=1$ and $\null m>n>0$.

So, $\null \gcd(2m,n)=1$ i.e $\null 2m=w^2$ and $\null n=r^2$ for some $\null m,w\in \mathbb{N}$. So, $\null m=2w_1^2$ for some $\null w_1$.

We have $\null x_0^2=(2w_1^2)^2+(r^2)^2$. So, $\null (2w_1^2,r^2,x_0)$ is a Pythagorean triple. So,

$\boxed{\null 2w_1^2=2gh,r^2=g^2-h^2, x_0=g^2+h^2,\gcd(g,h)=1,g\not\equiv h \bmod 2\dots *}$

That allows us to conclude that $\null g,h$ are perfect squares. $\null g= k^2$, $\null h=l^2$. So $\null r^2=k^4-l^4$ i.e. $\null (k,l,r)$ is a solution of the original equation.

Note that $\null 0 and this produces a contradiction by Fermat’s method of infinite descent.

Bibliography:

2. Elementary Number Theory} by David M. Burton(6th edition, McGraw Hill)

Here is a pdf of the above article: https://docs.google.com/file/d/0B4iuboCh7MplTWN6cFJ4bVV4dWc/edit

Please feel free to point out errors. A few may have inadvertently crept in due to the heavy editing involved(there were several LaTex issues- if you are having problems with typesetting math on wordpress, this webpage can be helpful: http://terrytao.wordpress.com/2009/02/10/wordpress-latex-bug-collection-drive/).