Let's write the following expression in a simpler form:

$$
\frac{(\sqrt{r+1}-\sqrt{r-1})^{n}-(\sqrt{r+1}+\sqrt{r-1})^{n}}{(\sqrt{r+1}-\sqrt{r-1})^{n}+(\sqrt{r+1}+\sqrt{r-1})^{n}}
$$

where $r=\left(x^{2}+y^{2}\right) / 2 x y, x$ and $y$ are different positive numbers, and $n$ is equal to 2, 3, or 4.
$$
K=\frac{(\sqrt{r+1}-\sqrt{r-1})^{n}-(\sqrt{r+1}+\sqrt{r-1})^{n}}{(\sqrt{r+1}-\sqrt{r-1})^{n}+(\sqrt{r+1}+\sqrt{r-1})^{n}}
$$

Substituting the given expression for $r$

$$
\begin{gathered}
\sqrt{r+1}=\sqrt{\frac{x^{2}+y^{2}}{2 x y}+1}=\sqrt{\frac{(x+y)^{2}}{2 x y}}=\frac{x+y}{\sqrt{2 x y}} \\
\sqrt{r-1}=\sqrt{\frac{(x+y)^{2}}{2 x y}}=\frac{|x-y|}{\sqrt{2 x y}}=\frac{x-y}{\sqrt{2 x y}}, \text { if } x>y(>0)
\end{gathered}
$$

Since the expression for $r$ is symmetric in $x$ and $y$, without loss of generality, we can assume that $x>y$; this ensures that $\sqrt{r-1} \geq 0$.

Substituting these into (1), in the numerator and denominator of $K$,

$$
\sqrt{r+1}-\sqrt{r-1}=\frac{2 y}{\sqrt{2 x y}}, \quad \sqrt{r+1}+\sqrt{r-1}=\frac{2 x}{\sqrt{2 x y}}
$$

and multiplying by $\left(\frac{\sqrt{2 x y}}{2}\right)^{n}$,

$$
K=\frac{y^{n}-x^{n}}{y^{n}+x^{n}}(<0)
$$

This is valid for all natural numbers $n$, not just the specified ones.

In the final form, the polynomials in the denominator and the numerator have no common polynomial factor, so further simplification is not possible. If they had a common factor, it would also be a factor of their sum, $2 y^{n}$, and their difference, $2 x^{n}$, but we see that this is not the case.

Remark. We did not fully utilize the assumptions about $x$ and $y$, only that they are non-zero and have the same sign.