A Solution in Natural Numbers
Date: 10/30/2001 at 06:25:18 From: Salman Subject: Number theory Hello Dr. Maths, Please help me in solving the following question: Prove that x^2+y^2 = z^n has a solution in 'natural numbers' for all 'n' where 'n' is a natural number. Thanks.
Date: 01/02/2002 at 15:53:56 From: Doctor Rob Subject: Re: Number theory Thanks for writing to Ask Dr. Math, Salman. Pick z to be any number of the form z = a^2 + b^2 where a > 0 and b > 0. Now prove your theorem by induction on n. Above is the case n = 1. Use the fact that the following identity holds: (t^2+u^2)*(v^2+w^2) = (t*v+u*w)^2 + (t*w-u*v)^2. Build up a representation of z^(n+1) from a representation of z^n as the sum of two squares, as the crucial part of the induction step. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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