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 . 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
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 and and the hypotenuse be of length , then . This motivates us to find all positive integral solutions of the form .
More specifically, we shall investigate equations of the form where .
Such triples are known as primitive Pythagorean triples.
Two lemmas are in order.
Lemma 1: In a Pythagorean triple,one and only one of and is even.
Proof:Note that by the definition of a primitive Pythagorean triplet. If both are even, . If both are odd, , a contradiction as any square is either or .So one of them is even, the other is odd.
Lemma 2: If and , then there exist positive integers such that .
: The proof is left to the reader.However, looking at the prime representation of and is helpful.
An Important Theorem
Theorem 1:The solutions to the equation where , , is given by
for some naturals and .
Proof: . Note that
If not, let . Then and .
Thus divides and . As , also divides which forces which is not true as their gcd is .
Using lemma 2,
and for some positive integers .
So, and , . Note that means which in turn implies that .Further, forces to be 1 as well.
QED.
Two More Theorems
Theorem 2: There is no solution to in positive integers .
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 .
Further, let . So there exist natural numbers such that so that , .
That implies for some in naturals.
We now have such that .
So we now have a Pythagorean triple and so we have (assuming is even) for some positive naturals with .
and are odd by lemma 1.
If is odd, , a contradiction ( note that if t is odd, s is even). So, is even.
Let . Then such that .
Consider the equation (note that implies that is 1. So, there exist natural numbers such that .
Once again, and for some natural numbers using the previous arguments.
So, . Note that this makes a solution set to our equation.
i.e for each , there is a smaller which is part of the solution of the equation. being finite, at some stage, a contradiction.This is the Fermat’s method of infinite descent.
So the equation has no solution in natural numbers.
Note that this implies that that there is no solution to the equation , the Fermat’s Last Theorem for the exponent .
Theorem 3: There is no solution to the equation in natural numbers.
Proof: Imitating theorem 1’s proof, we consider that case when for coprime .
So now there are two cases. If is odd,for some natural numbers , where and and .
Multiplying the first two equations, we get i.e. ,
: the Fermat’s Method of infinite descent shows that there is a contradiction!
We are left with the case is even.
Then for some natural numbers , where and .
So, i.e and for some . So, for some .
We have . So, is a Pythagorean triple. So,
That allows us to conclude that are perfect squares. , . So i.e. is a solution of the original equation.
Note that 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/).