Generalized Pythagorean Triples

$\text{Consider a primitive pythagorean triple }$ $(a,b,c)$ $\text{ such that}$

$\begin{aligned}a^2+b^2&=c^2\\ \text{multiplying throughout by } c^{2n} \text{ we get,}\\ (ac^n)^2+(bc^n)^2&=(c^2)^{n+1} \\ p^2+q^2&=r^{n+1}\\ \text{where } p=ac^n,q=bc^n \text{ and } r=c^2\end{aligned}\\ \text{since there are infinitely many Pythagorean triples and each one corresponds to a solution as shown above,}\\ \text{ there will be infinitely many solutions for each n.}$

One infinite family of solutions is $(a,b,c,n) = (2^{m}, 2^{m}, 2, 2m + 1)$ for any integer $m \ge 1$ .

If we define integers $a_{k,n},b_{k,n}$ by setting $a_{k,n} + ib_{k,n} \; = \; (2 + i)^{kn}$ then $a_{k,n}^2 + b_{k,n}^2 \; = \; \big|(2+i)^{2kn}\big| \; = \; 5^{kn}$ and so $\big(a_{k,n},b_{k,n},5^k,n\big)$ is a solution for any integers $k \ge 1$ and $n \ge 3$ . Thus we get infinitely many solutions for each value of $n \ge 3$ .

There are infinitely many integer solutions for $a^2+b^2=c^n$ , for any integer $n\geq2$ and $\{a,b,c\in\mathbb{Z}|a,b,c\neq0\}$ .

Note that I am not certain of the validity of this particular proof/formula. I will be grateful for any suggestions.

Consider any Gaussian integer $z$ . Perform the operation $z^n$ , and recall that $|z^n|^2=Re(z^n)^2+Im(z^n)^2,$ where $Re(z^n)$ and $Im(z^n)$ are respectively the real and imaginary coefficients of $z^n$ . Now, the complex norm has a nice multiplicative property: $|z^n|=|z|^{\frac{n}{2}}$ because $|z^n|=\sqrt{(Re(z^n)^2+Im(z^n)^2)^n}=(Re(z^n)^2+Im(z^n)^2)^{\frac{n}{2}}=(\sqrt{Re(z^n)^2+Im(z^n)^2})^n=|z|^n$ Substituting into the first equation, we obtain $|z|^{n}=Re(z^n)^2+Im(z^n)^2.$ As $Re(z^n)$ , $Im(z^n)$ , and $|z|$ are all positive integers, we have our proof.

Question: Does this formula include all possible integer solutions? What is the distribution of the solution amongst the integers?

An infinite family of solutions for $a^2 + b^2 = c^4$ is

$a = p^4 - 6p^2q^2 + q^4$

$b = 4p^3q - 4pq^3$

$c = p^2 + q^2$

for any integer $p$ and $q$ .

2^2 + 2^2 = 2^3

4^2 + 4^2 = 2^5

8^2 + 8^2 = 2^7

(2^x)^2 + (2^x)^2 = 2^(2x+1)