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Pythagorean Triples

Date: 07/14/97 at 08:09:27
From: Daniel Toscano
Subject: Pythagorean triples and extras

Is there a formula to determine the solutions to the following 

   a^2 + b^2 = c^2

(I have gained some info on this one by reading some of your 

Extending further:

   a^3 + b^3 + c^3 = d^3

   a^4 + b^4 + c^4 + d^4 = e^4

I am unable to work out solutions to these.  Can you help?

Note:  a^3 means - a to the power 3
       b^4 means - b to the power 4

Date: 07/14/97 at 08:47:04
From: Doctor Jerry
Subject: Re: Pythagorean triples and extras

Hi Daniel,

From The Theory of Numbers, by Hardy and Wright:

  "The most general solution of the equation x^2+y^2=z^2, satisfying 
   the conditions x>0, y>0, z>0, x and y have no common divisors, x is  
   divisible by 2 is x=2ab,  y=a^2-b^2,  and  z=a^2+b^2, where a, b  
   are integers of opposite parity and a and b have no common factors 
   and a,b>0. There is a 1-1 correspondence between different values 
   of a,b, and different values of x,y,z."

The Babylonians knew this theorem, more or less. Look at _The Exact 
Sciences in Antiquity_, by Otto Neugebauer, for more information on 
the Babylonians.

>a^3 + b^3 + c^3 = d^3
>a^4 + b^4 + c^4 + d^4 = e^4

I don't know the solutions to these equations. However, since it is 
now known that equations of the form x^k+y^k=z^k have no integer 
solutions for k>=3 (Fermat's Last Theorem), I would be surprised if 
there were solutions.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 07/25/97 at 07:58:05
From: Daniel Toscano
Subject: Advancing from Pythagorean triplets

To Doctor Math:

Can you please find values for the unknown variables in the two 
following equations:

w^3 + x^3 + y^3 = z^3

a^4 + b^4 + c^4 + d^4 = e^4

I know that it is possible for the variables to all have the value of 
zero, but I would like to know if there are any other possible values 
for the variables?  I am quite certain that for the second equation, 
one variable must have a value of zero while the other variables take 
on values in the thousands. I think the only way possible to discover 
any solutions is through trial and error - but it would be interesting 
to know if there is any mathematical way of discovering possible 
values.  Please help!


Date: 07/25/97 at 11:36:50
From: Doctor Rob
Subject: Re: Advancing from Pythagorean triplets

Several solutions are known for these equations.  For the first 
equation the smallest are:

     3^3 + 4^3 + 5^3 = 6^3
     1^3 + 6^3 + 8^3 = 9^3
   3^3 + 10^3 + 18^3 = 19^3
   7^3 + 14^3 + 17^3 = 20^3
   4^3 + 17^3 + 22^3 = 25^3
  18^3 + 19^3 + 21^3 = 28^3
  11^3 + 15^3 + 27^3 = 29^3

There are parametric formulae which give infinite sets of solutions, 
but not necessarily all solutions.  See Albert H. Beiler, _Recreations 
in the Theory of Numbers_, pp. 290-291.  

Another reference is G. H. Hardy and E. M. Wright, _The Theory of 
Numbers_, Section 13.7, pp. 199-201.

For the second equation, the smallest solution is:

  30^4 + 120^4 + 272^4 + 315^4 = 353^4.

An even harder problem is to make a = 0, and Leonhard Euler 
conjectured that this was impossible. He was wrong, and this was 
proved possible by Noam Elkies, who found the first example. Roger 
Frye found the smallest one:

  95800^4 + 217519^4 + 414560^4 = 422481^4.

Again a reference is Hardy and Wright, Section 21.11, pp. 332-335.  
No infinite families of solutions are known.

-Doctor Rob,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Number Theory

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