Prove that ${x^2+y^2=z^n}$ has a solution in $\mathbb{N}$ for all $n$ in $\mathbb{N}$

Question

I am solving it by stating that $$x^2 +y^2 =c^2$$ represents a circle. And when $$c^2=z^n$$ then , it represents a system of concentric circles with radius varying as $z$ varies or $n$ varies. So, for all $n$ in $\mathbb{N}$, $z^n$ represents the radius of the circle. Hence, $x^2 + y^2 =z^n$ has solution in $\mathbb{N}$.

This proof is not formal. It may even be wrong. I really need to know how to prove it formally using theorems from Number Theory. Please help me! Thank you! :)

Answer 1

Clearly $$ 5^2+10^2=5^3, $$ and hence $$ (5^{k+1})^2+(10\cdot 5^k)^2=5^{2k+3}. $$ Thus, for every $n$ odd your claim holds.

Then $$ 3^2+4^2=5^2, $$ implies that $$ (3\cdot 5^k)^2+(4\cdot 5^k)^2=5^{2+2k}, $$ Thus, for every $n$ even your claim holds.

Answer 2

Note

$$(2^k)^2+(2^k)^2=2\times 2^{2k}=2^{2k+1}$$

So there is a solution for ever odd n.

Answer 3

By Lagrange's identity: $$ (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2 $$ it follows that the numbers that can be represented as a sum of two squares form a semigroup.

Since every prime of the form $4k+1$ can be represented as a sum of two squares, a sufficient condition for $$x^2+y^2 = z^n $$ to be solvable over the integers is that $z$ has no prime divisors of the form $4k+3$.