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### A Solution in Natural Numbers

```
Date: 10/30/2001 at 06:25:18
From: Salman
Subject: Number theory

Hello Dr. Maths,

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/
```
Associated Topics:
High School Number Theory

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