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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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
Math is not really about opinions. If you think there are only four solutions, try and prove that every solution is one of those four. If you get stuck it might be a good start on a counterexample, or a proof that there are uncountably many solutions after all. If you succeed, then it's not an opinion anymore (beware of false proofs, though). – Asaf Karagila Jan 28 '13 at 14:19
up vote 5 down vote accepted

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.

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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 $$ x^2 = (ai + bj + ck)(ai + bj + ck) = -(a^2 + b^2 + c^2) = -1. $$

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