Replying to
@hallasurvivor This is a nice observation. Here is how I think about Puzzle 1: The unit quaternions are isomorphic to \( \mathbb{R}^3 \), and their product is given in terms of the dot product and cross product. Writing a general quaternion as \( (a, \vec{v}) \), with \( a \in \mathbb{R}, \vec{v} \in \mathbb{R}^3 \), the product can be written as (0,𝑣⃗)(0,𝑢⃗)=(−𝑣⃗⋅𝑢⃗,𝑣⃗×𝑢⃗).
So taking \( \vec{v} \in S^2 \) to be any unit imaginary quaternion, we find that its square is \( -1 \).