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"Prove that [itex]x^2 - 1=0[/itex]" has infinitely many solutions in the division ring Q of quaternions."

The Quaternions are presented in my book in the representation as two-by-two square matrices over ℂ. The book gives that for a quaternion

(sorry for the terrible notation, I wasn't able to figure out how to do a matrix in here and I didn't want to do the tex by hand)

[itex]\stackrel{a+bi\ \ c+di}{-c+di\ \ a-bi}[/itex]

has inverse

[itex]\frac{1}{a^2+b^2+c^2+d^2}\stackrel{a-bi\ \ -c-di}{c-di\ \ a+bi}[/itex].

Now if [itex]x^2-1=0[/itex], then clearly [itex]x=x^{-1}[/itex]. This means that the individual components must be equal, so a+bi is a real multiple of its conjugate, which requires that it is real, and that multiple is one, or that it is imaginary, and that multiple is -1. Since the multiple is identically positive (sum of squares) b=d=0, and since the multiple must equal 1, [itex]a^2+c^2=1[/itex]. This gives us

[itex]\stackrel{a\ \ c}{-c\ \ a}=\stackrel{a\ \ -c}{c\ \ a}[/itex],

so c=-c=0. Therefore, a=1, which seems to imply that 1 (the quaternion unity) is the unique solution.

What am I missing?

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# Quaternion Polynomial Equation

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