While studying the notions of Extension fields and Splitting fields, I came across to handle the following question:
Consider the real quaternion ring $R$. Then for any $t \in R$ satisfying $t^2=-1$, prove that we can find a $s \in R$ such that $sts^{-1} = i$.
First I have found that, every $t \in R$ being a square of $-1$ in $R$ is pure imaginary unit quaternion; that is \begin{gather*} t = a_1 i + a_2 j + a_3 k, \quad a_1^2 + a_2^2 + a_3^2 =1 \end{gather*} for any real $a_1, a_2, a_3 \in \mathbb{R}$. And I also have found that we can isomorphically identify the real quaternions with the elements in $M_2(\mathbb{C})$ as \begin{gather*} a_0 + a_1 i + a_2 j + a_3 k \iff \begin{pmatrix} a_0+a_1i & a_2+a_3i \\ -a_2+a_3i & a_0-a_1i\end{pmatrix}. \end{gather*} Then I merely tried to make a matrix calculation if what kind of element would $s^{-1}is$ be. By identifying $i \in R$ with $\begin{pmatrix} i &0 \\ 0 & -i\end{pmatrix}$, I came to conclude that $t$ satisfying $t=s^{-1}is$ must be $-i$. But, this is definitely not the question wanted. So I have some doubts; whether I am misunderstanding the question or there is another brilliant way to tackle the problem.
If it was to find some $t$ (such that $t^2=-1$) and $s \in R$ satisfying $sts^{-1} =i$, the problem would be too easy. I like to get some advice. Please help!
It was a good start that you figured out that we must have $$ t=a_1i+a_2j+a_3k $$ for some triple $(a_1,a_2,a_3)\in\Bbb{R}^3$ such that $a_1^2+a_2^2+a_3^2=1$. To find a suitable $s$ we do the following.
Case 1. If $a_2=a_3=0$ the task is easy. We must have $t=\pm i$. If $t=i$ we can use $s=1$. If $t=-i$ we can use $s=j$ as $jij^{-1}=-i$.
Case 2. Either $a_2\neq0$ or $a_3\neq0$. In this case the vectors $i$ and $t$ are linearly independent. Therefore their cross product $i\times t$ is a non-zero vector. Furthermore, the angle $\theta$ between $i$ and $t$ is in the interval $\theta\in(0,\pi)$. Here we can use the following trick (well known to people using quaternions when implementing 3D-rotations in e.g. computer graphics, see also runway44's comment). Let $u$ be the vector $i\times t$ scaled to have unit length. Then $$ s=\cos(\theta/2)-u \sin(\theta/2) $$ will work. This is because the linear transformation $v\mapsto svs^{-1}$ is the rotation about the axis $u$ by angle $\theta$. Remark: More often the sources uses $$ s^{-1}=\cos(\theta/2)+u\sin(\theta/2), $$ but then the rotation would map $i$ to $t$ rather than the other way arouns as we want here.
Example. Let $t=j$. We know that $i\times j=k$, so $u=k$. We know also that the angle between $i$ and $j$ is $\theta=\pi/2$. The recipe gives $$ s=\cos\frac{\pi}4-k\sin\frac{\pi}4=\frac{1-k}{\sqrt2}. $$ This has unit norm, so we know that $$ s^{-1}=\overline{s}=\frac{1+k}{\sqrt2}. $$ We can verify that $$ \begin{aligned} sjs^{-1}&=\frac1{\sqrt2}(1-k)j\frac1{\sqrt2}(1+k)\\ &=\frac12(j-kj)(1+k)\\ &=\frac12(j+i)(1+k)\\ &=\frac12(j+jk+i+ik)\\ &=\frac12(j+i+i-j)\\ &=i \end{aligned} $$ as prescribed.