# Do there exist equations that cannot be solved in $\mathbb{C}$, but can be solved in $\mathbb{H}$?

Excluding polynomials (whose solutions are covered by the Fundamental Theorem of Algebra), do there exist any univariable equations that cannot be solved in the complex numbers, but can be solved other fields, such as the quaternions, octonions, etc.?

I know that not all equations can be solved (in any field), such as $e^w=0,$ or $0x=0$ (no unique solution), and also that some equations that can be solved in $\mathbb{C}$ can also be solved in $\mathbb{H}$, but with many more solutions (e.g. $w^2+1=0$), but I'm wondering whether there exist equations that can only be solved in higher-dimensional division algebras (than $\mathbb{C}$)?

Thanks

• Do you mean a division algebra instead of "higher dimensional fields"? $\mathbb{H}$ is not commutative. – Bombyx mori May 10 '14 at 19:39
• Oh, yes! So I suppose I should remove sedenions from this list, then. So just $\mathbb{H}$ and $\mathbb{O}$. – beep-boop May 10 '14 at 19:41
• $(\textrm{i} \, x-x\, \textrm{i})^2+1=0$ – WimC May 10 '14 at 19:44
• Ah. I was wondering why Bombyx mentioned that $\mathbb{H}$ is not commutative. But it makes sense in this case, that this equation couldn't be solved in $\mathbb{C}$. Thanks! – beep-boop May 10 '14 at 19:47
• @alexqwx If you can't think of a more clever way, you can always expand $x$ into $a+bi+cj+dk$ and multiply. Hastily doing some computations, I think I found that anything with $c^2+d^2=\frac14$ would do the trick. – rschwieb May 13 '14 at 1:59

$(\textrm{i} \, x - x \, \textrm{i})^2+1=0$