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.


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<m_1^2=m<m^2+n^2=s<2st<x_0^2\implies 0<m_1<x_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<m<\sqrt{m^2+n^2}=x_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<k=\sqrt{g}<g^2+h^2=x_0 and this produces a contradiction by Fermat’s method of infinite descent.


1.  Fermat’s Last Theorem for Specific Exponents

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

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