# $x^2+1=0$ uncountable many solutions [duplicate]

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

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## marked as duplicate by MJD, Seirios, Javier Álvarez, Davide Giraudo, Stefan HansenJan 28 '13 at 14:56

If $a^2+b^2+c^2=1$ then $ai+bj+ck$ is solution of this equation. – Hanul Jeon Jan 28 '13 at 14:18
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.
Thanks, this seems clear to me. Was it intuition considering numbers of the form $x = \sqrt p i + \sqrt{1-p} j$$– Voyage Jan 28 '13 at 14:18 @Voyage I first tried the form pi + (1-p)j, and when I saw where that went wrong it was clear that taking square roots would make it work. – Jonathan Christensen Jan 28 '13 at 14:19 Let a, b, c be real numbers satisfying a^2 + b^2 + c^2 = 1 and let x = ai + bj + ck. Then$$ x^2 = (ai + bj + ck)(ai + bj + ck) = -(a^2 + b^2 + c^2) = -1.$\$