If x^2 is in F, x not in F, then x is a Pure Quaternion

Proposition: If $x^2\in Z(A)=F$ and $x\notin F$ , then $x$ is a pure quaternion.

Proof: Let $x=c+z$ , with $c\in F$ and $z\in A_+$ ( $z$ is a pure quaternion).

Then $x^2=c^2+2cz+z^2=c^2-\nu(z)+2cz$ . The key observation is that if $z=c_1i+c_2j+c_3k$ , then $z^2=-\nu(z)=ac_1^2+bc_2^2-abc_3^2$ .

Since $x\in F$ , this means that $2cz=0$ , i.e. $c=0$ , so $x$ is a pure quaternion.

Author: mathtuition88