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Beyond Pythagorean Triples


Date: 11/20/2001 at 10:49:52
From: lynn meredith
Subject: Pythagorean triples

I am a pupil studying a topic that covers the Pythagorean triples. 
For the higher piece I  need to see and prove whether you can use the 
equations:

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

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

Could you perhaps show me how these work, how you obtain numbers, and 
how to prove they are correct?

Thank you!


Date: 11/20/2001 at 12:59:53
From: Doctor Rob
Subject: Re: Pythagorean triples

Thanks for writing to Ask Dr. Math, Lynn.

Both of these equations are much more difficult than the situation
with Pythagorean triples.

Albert H. Beiler, _Recreations in the Theory of Numbers - The Queen
of Mathematics Entertains_ (New York: Dover Publications, 1964),
says, on page 220,

 "The more general relation x^3 + y^3 + z^3 = t^3 also has an
  infinitude of solutions, the most simple of which are
  3^3 + 4^3 + 5^3 = 6^3  and  1^3 + 6^3 + 8^3 = 9^3.

  There is a fairly complicated formula for all the solutions of
  this equation, but the formula for all those of a certain type
  is simple enough to find a place here. It is the identity:
  a^3(a^3+b^3)^3 = b^3(a^3+b^3)^3+a^3(a^3-b^3)^3+b^3(2a^3-b^3)^3.

  When a = 2, b = 1, this gives 18^3 = 9^3 + 12^3 + 15^3, which
  reduces to 6^3 = 3^3 + 4^3 + 5^3. When a = 3, b = 1, we obtain
  84^3 = 28^3 + 53^3 + 75^3, and for a = 3, b = 2, we have
  105^3 = 33^3 + 70^3 + 92^3.

  Another identity is
  a^3(a^3+2b^3)^3 = a^3(a^3-b^3)^3+b^3(a^3-b^3)^3+b^3(2a^3+b^2).

  Here for a = 2, b = 1, we have 20^3 = 7^3 + 14^3 + 17^3, and for
  a = 3, b = 1 there results 87^3 = 26^3 + 55^3 + 78^3.
  Another relation yields 11^3 + 15^3 + 27^3 = 29^3."

The equation a^4 + b^4 + c^4 = d^4 was conjectured by Leonhard Euler 
in the 18th century to have no solutions. No proof was forthcoming, 
nor were any solutions found until just a few years ago, when Noam 
Elkies found a fairly large solution, using the theory of elliptic 
curves. The smallest one was produced by Roger Frye using a computer 
search:

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

No general formulae are known, nor is it known whether or not the
number of solutions is infinite (although that is what I would
guess).

Feel free to write again if I can help further.

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

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