Complex equations with no complex solutions? Are there complex equations that admit no complex solutions, but rather quaternions or hypercomplex solutions, for example, in complete analogy to, say, the equation $x \times x = -1$ when restricted to the real line?
Edit: I am indeed using "equation" in its broader sense, as I am not restricting it to operations involving exclusively complex multiplications, say.
 A: Using the definition of the quaternions,
$$x^2=y^2=z^2=xyz=-1\quad\quad\quad x,y,z\in\mathbb{C},\quad (x\neq y)\wedge(x\neq z)\wedge(y\neq z)$$
The complex solutions to $x^2=-1$ are $x=i$ and $x=-i$ only. The equation needs a third solution but no more exist so $x$, $y$ and $z$ can't all be complex numbers at the same time. However, by definition, there are quaternion solutions.
A: Simple example, find a nonzero solution to $x^2=0$. This is easy in the dual numbers $\mathcal{N} = \mathbb{R}\oplus \epsilon \mathbb{R}$ where $\epsilon^2=0$. Indeed, the equation $x^2=0$ is solved by any element of $\epsilon \mathbb{R}$ as $x=\epsilon r$ for $r \in \mathbb{R}$ has $x^2=(\epsilon r)^2 = (\epsilon r)(\epsilon r) = \epsilon^2r^2=0$.
For hyperbolic numbers, you can find zero divisors. For quaternions, you can find infinitely many solutions to a quadratic equation. The distinction between nonstandard number systems and fields is quite noticeable. 
A: The notion of "equation" is very broad… 
I guess what you are looking for is the theorem telling you that every polynomial equation with complex coefficients always has a complex solution.
So you will never need quaternions or whatever for those equations.
