TITLE: $x^2+1=0$ uncountable many solutions
QUESTION [13 upvotes]: Possible Duplicate:
Why are the solutions of polynomial equations so unconstrained over the quaternions? 

Coudl someone explain me the following: Why should $x^2+1=0$ have uncountable infinite many solutions $x\in\mathbb H$?
In my opinion it has only 4 solutions, namely $i^2=j^2=k^2=ijk=-1$ ?

REPLY [13 votes]: Let $a, b, c$ be real numbers satisfying $a^2 + b^2 + c^2 = 1$ and let $x = ai + bj + ck$. Then
$$
x^2 = (ai + bj + ck)(ai + bj + ck) = -(a^2 + b^2 + c^2) = -1.
$$