It was an amazing one month at Goa University. In case you are wondering, I went there to attend level O of the Mathematics Training and Talent Search Programme. The schedule was somewhat hectic with a few breaks in between. Classes would start at 9 am and end at 5 pm. The instructors were available even after the class-timings and students and teachers would have breakfast, lunch and dinner together everyday. The programme emphasized independent thinking and was designed to discourage students from looking at the solutions directly without any attempt. There were lectures in logic and foundations, linear algebra, analysis and number theory along with group discussions.The most interesting bit, in my opinion, was the set of students’ seminars. In my case, I decided to talk about how in ordered field F, the least upper bound axiom and the greatest lower bound axiom are equivalent.

And of course, we did visit a few places as well. We ended up going to a few churches, beaches, Aguada fort and a few other temples. The focus however was on mathematics. The teaching was amazing! I still remember how Dr Jayanthan motivated the proof of the Chinese Remainder Theory when the option of simply supplying a solution to the system of linear congruences was so simpler.

If there’s one word to describe the programme, it would be “amazing”. Indeed, I had a good time in Goa.

Advertisements

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<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.

Bibliography:

1.  Fermat’s Last Theorem for Specific Exponents

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/).