# Quaternion Polynomial Equation

by alexfloo
Tags: equation, polynomial, quaternion
 P: 192 I'm working on the following: "Prove that $x^2 - 1=0$" 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) $\stackrel{a+bi\ \ c+di}{-c+di\ \ a-bi}$ has inverse $\frac{1}{a^2+b^2+c^2+d^2}\stackrel{a-bi\ \ -c-di}{c-di\ \ a+bi}$. Now if $x^2-1=0$, then clearly $x=x^{-1}$. 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, $a^2+c^2=1$. This gives us $\stackrel{a\ \ c}{-c\ \ a}=\stackrel{a\ \ -c}{c\ \ a}$, so c=-c=0. Therefore, a=1, which seems to imply that 1 (the quaternion unity) is the unique solution. What am I missing?
 Sci Advisor P: 1,546 Are you sure there isn't a typo in the question? I would expect $x^2 + 1 = 0$ to have infinitely many solutions. $x^2 - 1 = 0$ should have two solutions.
 Sci Advisor P: 906 if q* is the quaternial conjugate, then qq* is real. note that q-1 = q*/|q|, so from q = q-1, we get q = q*/|q|, implying that q is real. but x2 -1 = 0 has just two real solutions, 1 and -1. if it IS a typo, consider (ai + bj + ck)2, where a2+b2+c2 = 1 (which has infinitely many solutions (a,b,c)).
P: 192

## Quaternion Polynomial Equation

In fact, it wasn't a typo at all. The problem just before it references the polynomial $x^2-1$ over an entirely different field, and I misread. Apologies, and thanks for your help!

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