Prove that $x² + y² = z^n$ has a solution in $\mathbb{N}$, for all $n$ belonging to the set of natural numbers

Question

Prove that $x^2 + y^2 = z^n$ has a solution in $\mathbb{N}$, for all $n$ belonging to the set of natural numbers.

I think it uses Induction as a method of proof but just can't proceed with three variables. Any help is greatly appreciated.

Answer 1

Proof by induction:

base case: $n = 1$

$x^2 + y^2 = z$

for any integers $x,y, x^2 + y^2$ is an integer

Suppose for all $k\le n$ there exists integers $x,y,z$ such that

$a^2 + b^2 = z^k$ and $p^2 + q^2 = z$

We must show that there exists integers $x,y,z$ such that:

$x^2 + y^2 = z^{n+1}$

$z^{n+1} = z(z^n) = (p^2 + q^2)(a^2 + b^2)\\ (ap)^2 + (aq)^2 + (bp)^2 + (bq)^2\\ (ap)^2 + (aq)^2 + (bp)^2 + (bq)^2 + 2abpq - 2abpq\\ (ap + bq)^2 + (bp - aq)^2 = z^{n+1}$