Solutions of the equation $x^2+1 = 0$

Question

Let's consider the equation $$x^2+1 = 0$$

Now, if we work in $\mathbb{R}$, there are no solutions.

If we work in $\mathbb{C}$, there are two solutions.

Now I have been studying rings and modular algebra, and I understood that is we work in $\mathbb{Z}/65\mathbb{Z}$ there are actually four solutions.

A friend of mine replied telling me "working with quaternions, and there are infinite solutions".

Now I am not familiar with quaternions but I took a look and understood that those numbers satisty those relations

$$i^2 = j^2 = k^2 = -1 \qquad \qquad ij = jk = ki = 0$$

I now am wondering why there are infinite solutions. I mean, maybe is a very simple reasoning but correct me if I am wrong: I could simply take

$$x^2 = i^2 + \alpha jk$$

for any $\alpha$ to solve the equaton? Clearly since there are infinite good values for $\alpha$, I get infinite solutions.

Is that simple?

Answer 1

Let $a, b, c$ $\in\mathbb{R}$, satisfying $a^2 + b^2 + c^2 = 1$ and let $x = ai + bj + ck$.

Then

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

P.s. the properties you wrote are not correct.

Answer 2

First let me review a geometric intepretation of the equation $x^2=-1$ in the complex plane. We can visualize the whole complex plane $\mathbb C$ pretty easily using "real--imaginary" coordinates, and then we can visualize its "imaginary axis" $\{b i \mid b \in \mathbb R\}$. In that axis, by setting $x=bi$ with $|b| = 1$ we get two solutions $b=-1$ and $b=+1$ to the equation $x^2=-1$.

Although it's hard to visualize the whole of the 4-dimensional space of quaternions, we can nonetheless visualize the whole of its "imaginary subspace", namely the 3-dimensional vector space $$\{bi + cj + dk \mid b,c,d \in \mathbb R\} $$ In that subspace, the set of points $x = bi + cj + dk$ with $|(b,c,d)| = \sqrt{b^2+c^2+d^2}=1$ is infinite, in fact it is the entire unit 2-sphere in 3-space. And now I'm sure you can do the calculation yourself (using the corrected formulas for $ij$, $jk$, etc. in the comments): $$x^2 = (bi + cj + dk)^2 = -1 $$