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I am working through Kevin Houston's book How to think like a mathematician. The question I am currently stuck on is to show that

$$x^2 + y^2 = z^n$$

has positive integer solutions for $n = 1,2,3,\dots$

My work thus far:

$f: \mathbb{N}^2 \rightarrow \mathbb{N}$ defined by $f(x,y) = x^2 + y^2$ with image $A = \{2,5,8,10,13,17,\dots\}$

$g: \mathbb{N} \rightarrow \mathbb{N}$ defined by $g(z) = z^n$ with image $B = \{1,2^n,3^n,\dots\}$

If I can show that $A \cap B \neq \varnothing$ for $n \in \mathbb{N}$ then this shows that there are integer solutions for all $n$. How can this be shown?

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  • How is this a set theory question? – Asaf Karagila Oct 07 '17 at 13:35
  • Hint for getting started on half the problem. If $n$ is even then right triangles will help. – Ethan Bolker Oct 07 '17 at 13:36
  • @AsafKaragila maybe they thought, hey $z^n$ needs to be in the set of numbers that are the sum of two squares ? –  Oct 07 '17 at 13:39
  • Sorry for going a bit off-topic but this is a very well-known problem which is easily proved using Gaussian integers. Consider $x+iy\in\Bbb Z[i]$ such that $x^2+y^2=|x+iy|^2=z$ where $z$ is a positive integer. The existence of such a number is obvious. Now, notice that for any two gaussian integers, we have $|(a+ib)(c+id)|^2=(ac-bd)^2+(bc+ad)^2$ which is still a sum of two squares. So, for any gaussian integer $\bf x$, we note that $|\bf x|^2$ is a sum of two squares which immediately gives us a way to generate solutions for all $n\geq 2$ with just the solution for $n=1$. – Prasun Biswas Oct 07 '17 at 13:40
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    @PrasunBiswas: Expand your comment and post it as an answer. – MrYouMath Oct 07 '17 at 13:41
  • @MrYouMath, I don't think it would be a relevant answer to a question marked with the "set theory" tag. – Prasun Biswas Oct 07 '17 at 13:43
  • Tags: Maybe the OP mistook elementary set theory for elementary number theory. – Oscar Lanzi Oct 07 '17 at 13:53
  • It seems that my approach based on the intersection of the image sets may be not be plausible. This is probably not a problem solved using set theory and the other question contains the answer. – egg Oct 07 '17 at 13:56
  • for all odd powers it's as simple as as x and y being the same power of two. you then have $2^{2a}+2^{2a}=2^{2a+1}$ for $a$ a natural number. this covers all odd powers. –  Oct 07 '17 at 14:05

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