Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Proving Fermat's Last Theorem for N = 4


Date: 05/18/2000 at 09:03:55
From: Henryk Dabrowski
Subject: Fermat's Last Theorem

I don't know how to prove Fermat's Last Theorem for the specific case
n = 4. Can you help me?


Date: 05/18/2000 at 14:51:07
From: Doctor Rob
Subject: Re: Fermat's Last Theorem

Thanks for writing to Ask Dr. Math, Henryk.

To show that x^4 + y^4 = z^4 is impossible, we actually prove the 
stronger fact that x^4 + y^4 = w^2 is impossible, whether or not w is 
a perfect square, except when x*y = 0.

First we restrict our attention (without loss of generality) to the 
case GCD(x,y) = 1. Assume that there is a solution (x,y,w) with x > 0 
and y > 0. Among all of them, pick the one with the smallest w > 0 
(this exists because every nonempty subset of the natural numbers has 
a smallest element).

Then (x^2,y^2,w) is a primitive Pythagorean triple. Assume (without 
loss of generality) that y^2 is the even member of the triple. Then 
there exist two integers u and v with:

     u > v > 0,
     GCD(u,v) = 1,  and 
     u + v odd,

such that:

     x^2 = u^2 - v^2,
     y^2 = 2*u*v,  and
     w = u^2 + v^2.

If u were even, then v would be odd, and x^2 = 0 - 1 = 3 (mod 4), 
which is impossible. Thus u is odd and v is even. Then

     (y/2)^2 = u*(v/2),
     GCD(u,v/2) = 1.

This implies that for some integers r and s,

     u = r^2,
     v/2 = s^2,
     y = 2*r*s,
     GCD(r,s) = 1,
     r > 0, s > 0,  and
     r is odd.

Also x^2 + v^2 = u^2, so we have x^2 + 4*s^4 = r^4. Since 
GCD(r,2*s) = 1, we have another primitive Pythagorean triple 
(x,2*s^2,r^2). Thus there are integers a and b such that

     x = a^2 - b^2,
     2*s^2 = 2*a*b,
     r^2 = a^2 + b^2,
     GCD(a,b) = 1,
     a > b > 0,  and
     a + b odd.

Since a*b = s^2, we can write a = f^2, b = g^2 with some f > 0 and
g > 0 with GCD(f,g) = 1. Then we have r^2 = f^4 + g^4, which looks 
just like our original equation. Now

     w = u^2 + v^2 = r^4 + 4*s^4 > r^4 >= r > 0,

so w > r > 0 and (f,g,r) is a different solution from (x,y,w), and its 
last component is smaller. This is a contradiction. Thus our 
assumption that a solution (x,y,w) exists with x > 0 and y > 0 must be 
incorrect. This means that the equation x^4 + y^4 = w^2 has no integer 
solution unless x*y = 0. This implies Fermat's Last Theorem for the 
case n = 4.

This is a classic application of Fermat's Method of Infinite Descent. 
In fact, it is a proof essentially due to Fermat himself. I have 
adapted the proof from Ivan Niven and Herbert S. Zuckerman, _An 
Introduction to the Theory of Numbers_ (1960), pp. 100-102.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Number Theory
High School Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/