Pythagorean TriplesDate: 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 equations? a^2 + b^2 = c^2 (I have gained some info on this one by reading some of your articles.) 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! http://mathforum.org/dr.math/ 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! Daniel 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! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/