A pair of continued fractions that are algebraic numbers and related to $a^2+b^2=c^m$ Similar to the cfracs in this post, define the two complementary continued fractions,
$$x=\cfrac{-(m+1)}{km\color{blue}+\cfrac{(-1)(2m+1)} {3km\color{blue}+\cfrac{(m-1)(3m+1)}{5km\color{blue} +\cfrac{(2m-1)(4m+1)}{7km\color{blue}+\cfrac{(3m-1)(5m+1)}{9km\color{blue}+\ddots}}}}}\tag1$$
$$y=\cfrac{-(m+1)}{km\color{red}-\cfrac{(-1)(2m+1)} {3km\color{red}-\cfrac{(m-1)(3m+1)}{5km\color{red}-\cfrac{(2m-1)(4m+1)}{7km\color{red}-\cfrac{(3m-1)(5m+1)}{9km\color{red}-\ddots}}}}}\tag2$$
The first one is the superfamily which contains Nicco's cfracs in another post. Let $i$ be the imaginary unit. For $k>1$ and $m>1$, it can be empirically observed that $x$ obeys,
$$\left(\frac{(x+i)^m-(x-i)^m}{(x+i)^m+(x-i)^m}\right) \color{blue}{\left(\frac{(k+i)^{m+1}+(k-i)^{m+1}}{(k+i)^{m+1}-(k-i)^{m+1}}\right)^{(-1)^m}}=1\tag3$$
while $y$ obeys,
$$\left(\frac{(y+1)^m+(y-1)^m}{(y+1)^m-(y-1)^m}\right) \color{blue}{\left(\frac{(k+1)^{m+1}+(k-1)^{m+1}}{(k+1)^{m+1}-(k-1)^{m+1}}\right)^{(-1)^{m+1}}}=-1\tag4$$
where the colored part is a constant that depends on the choice of $k,m$. Hence, as shown in this post, $x,y$ are radicals and algebraic numbers of degree $m$.

Question: How do we prove that $(3)$ and $(4)$ are indeed true?

P.S. Since,
$$\left(\frac{(z+i)^m+(z-i)^m}{2}\right)^2+i^2\left(\frac{(z+i)^m-(z-i)^m}{2}\right)^2 = (z^2+1)^m$$
then the structure of $(3)$ explains the observations about $a^2+b^2=c^m$ in Nicco's post.

Using the general solution from this post,we obtain
$$x=\dfrac{ik\sqrt[m]{\dfrac{k+i}{k-i}}-\sqrt[m]{\dfrac{k+i}{k-i}}-ik-1}{k\sqrt[m]{\dfrac{k+i}{k-i}}+i\sqrt[m]{\dfrac{k+i}{k-i}}+k-i}$$
$$y=\dfrac{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}-k-1}{k\sqrt[m]{\dfrac{k-1}{k+1}}-\sqrt[m]{\dfrac{k-1}{k+1}}+k+1}$$
For $m\gt2$
