TITLE: Does my definition of double complex noncommutative numbers make any sense?
QUESTION [22 upvotes]: I wanted to factorize $a^2+b^2+c^2$ into two factors in a similar way to $$a^2+b^2 = (a+ib)(a-ib)$$
This doesn't seem to be possible using real or complex numbers. However I came up with the following idea $$ (a + ib + jc) (a -ib -jc) = a^2+b^2+c^2$$ if we define $$i^2=j^2=-1$$ and $$ ij = -ji.$$ So instead of using complex numbers I had to define a kind of double complex number obeying a non-commutative multiplication rule to solve my factorization problem.
Does my definition of this numbers make sense or is it somehow inconsistent? Do these numbers already have a name in mathematics and are they used? Is there any literature about them?

REPLY [3 votes]: It's worth pointing out that you need to go up to the quaternions - a four-dimensional space - because a factorization of the sort you're describing doesn't 'work' in just three dimensions.  Suppose you had $z=a+bi+cj$ (and so $\bar{z}=a-bi-cj$) and $w=d+ei+fj$ (with $\bar{w}=d-ei-fj$).  Then $x=zw$ would be of the form $r+si+tj$, with $r$, $s$ and $t$ each expressions in $a\ldots f$, and $\bar{x}$ would be $r-si-tj$; taking the norms (or in other words looking at $|x|^2 = x\bar{x} = |z|^2|w|^2$) would give you an identity of the form $(a^2+b^2+c^2)\cdot(d^2+e^2+f^2) = r^2+s^2+t^2$.  But consider $(1^2+1^2+1^2)\cdot(4^2+2^2+1^2) = 3\cdot 21 = 63$; this number is of the form $8n+7$ and so by the Sum Of Three Squares theorem it can't be expressed as $r^2+s^2+t^2$ for any values of $r,s,t$.  This means the three-squares identity (or in other words, the three-dimensional product you're looking for) can't exist.