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Show that the quaternion division ring contains an infinite number of elements $ u $ satisfying $ u^2=-1 $

I was trying to solve the above exercise on Page 133, Basic Algebra, Jacobson. Maybe it is convenient to consider $$ \mathbb H=\left\{ \left[\begin{matrix} \alpha & \beta\\-\bar{\beta} & \bar{\alpha} \end{matrix} \right]: \alpha, \beta\in\mathbb C \right\}. $$

Of course, $$ i= \begin{bmatrix} \sqrt{-1} & 0\\ 0 & -\sqrt{-1} \end{bmatrix},\quad j=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},\quad k=\begin{bmatrix} 0 & \sqrt{-1}\\ \sqrt{-1} & 0 \end{bmatrix} $$ satisfy $ u^2=-1 $. But how to show that there are infinitely many elements in $ \mathbb H $ satisfying the identity?

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    $\begingroup$ "Maybe it is convenient....". It isn't. $\endgroup$ Jan 13, 2019 at 8:49

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Hint: Forget about the matrices for a moment. What is $(ai+bj)^2$?

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The proof of this statement can be found on Wikipedia. Let the square root of $-1$ be $x=a+bi+cj+dk$, then $x^2=-1$ is equivalent to $$a^2-b^2-c^2-d^2=-1\qquad2ab=2ac=2ad=0$$ If $a\ne0$, $b=c=d=0$ and the first equation becomes $a^2=-1$, impossible since $a\in\mathbb R$. Hence $a=0$, at which point the second set of equations is automatically satisfied and the first equation becomes $b^2+c^2+d^2=1$. The quaternionic square roots of $-1$ thus form the unit sphere in $\mathbb R^3$, so there are infinitely many solutions.

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Use the idea you mentioned. Note that $$\begin{pmatrix}-1&0\\0&-1\end{pmatrix} =\begin{pmatrix}0&\beta\\-\bar\beta&0 \end{pmatrix}^2$$ for any $\beta\in\mathbb C$ with absolute value 1. There are, of course, infinitely many such matrices.

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