Let $x, y \in \mathbb{R}^{+}$ be such that $x \neq y$ and assume $n \in \mathbb{N}-\{0\}$.
Now, referring to the well-known hyperoperation sequence $x[n]y$, we have that $x[1]y=x+y$ and we know that addition is commutative.
If $n=2$, then we observe the same thing, since $x[2]y=x \cdot y$ and multiplication is also commutative, but $x[n]y$ is not commutative anymore as long as $n \geq 3$.
In particular, $n=3 \Rightarrow x=\nu^{\frac{1}{\nu-1}} \wedge y=\nu^{\frac{\nu}{\nu-1}}$, where we have set $y=\nu \cdot x$ (see Goldbach's solution provided in 1729).
Question: What happens if $n=4$ so that we need to solve the tetrational equation $^y{x} =$ $^x{y}$ (i.e., $x[4]y=y[4]x$)?
In general, is it (theoretically) possible to find a general solution to $x[n]y=y[n]x$, for any arbitrarily large (integer) value of $n$?