Is there such a thing as an equation with noncomplex quaternion solutions?

Question

I'm familiar with equations with real solutions and equations with nonreal complex solutions. Examples:

$x^2-3x+1=0$

has the real solutions

$3\pm \sqrt{5} \over 2$

and this other equation:

$3x^2-x+2=0$

has the nonreal complex solutions

$1\pm i\sqrt{23} \over 6$

but to my understanding, complex numbers are just a special kind of quaternion numbers, so I wonder if there's such a thing as an equation (not necessarily a polynomial equation) with noncomplex quaternion solutions. If there's such a thing, I would like to see an example.

Answer 1

Complex numbers have a very special property, they are an algebraically closed field. This means that any polynomial has complex solutions. If $P=a_nx^n+\dots +a_1x+a_0$ is a polynomial with complex coefficients then there exist $n$ complex numbers $\alpha,r_1,r_2\dots r_n$ (not necessarily distinct) such that:

$a_nx^n+\dots a_0=\alpha(x-r_1)(x-r_2)\dots (x-r_n)$.

So suppose $z$ is a non-complex quaternion root of $P$. Then we would have:

$(z-r_1)(z-r_2)\dots (z-r_n)=0$. This is impossible, the quaternions are a domain, this means that if a product of quaternions is zero then one of the factor must be zero. So we have $z-r_k=0$ for some $k$, implying $z=r_k$, meaning $z$ is complex.

Answer 2

Let's solve $ixi=x$. Explicitly, if $x=a+bi+cj+dk$, the left side is $-a-bi+cj+dk$. So this equation's solutions are those numbers of the form $cj+dk$ - those quaternions with no "complex part".