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.

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