$x^2+1=0$ uncountable many solutions

Question

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 $i^2=j^2=k^2=ijk=-1$ ?

Answer 1

Let $0 \leq p \leq 1$, and consider numbers of the form $$x = \sqrt p i + \sqrt{1-p} j$$ Taking the square, we have $$x^2 = (\sqrt p i + \sqrt{1-p} j)^2 = pi^2 + (1-p)j^2 + \sqrt{p(1-p)}(ij + ji)$$ The third term is zero, since $ij = -ji$. The first two terms sum to -1, so any number of the given form is a solution to $x^2 + 1 = 0$. Since there are uncountably many $p \in [0,1]$, there are uncountably many solutions.

Answer 2

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