Split-complex numbers and their possible application Suppose that there is number $a+jb$ where $j^2=1$ and the whole number is split-complex number. We want to set this number to satisfy the following: 
A) $(a+jb)(a+jb) = k(c+jd)$ where $k$ is fixed integer
B) for any natural number $z$, there exists $z$ split-complex numbers that satisfy A) with same $k$, but for two such different numbers, $(a+jb)(c+jd) \neq k(e+jf)$ where $e+jf$ is any number.
Is this possible?
 A: Let $s:=1+j$. Then $s^2=2+2j=2s$. So, for a $t\in\Bbb R$, we have
$$(t\cdot s)^2=t^2\cdot 2s=2t\,(t\cdot s)\,.$$
This answers A): for a fixed $k$, let $t:=k/2$, and consider $t\cdot s$, i.e. $\displaystyle\frac{k+kj}2$.
For B), let's look for all eigenvectors of this "split-complex" squaring. We already know that for $u:=s/2=\displaystyle\frac{1+j}2$ we have $u^2=u$, and similarly, $v:=\displaystyle\frac{1-j}2$, that gives $v^2=v$. We also have 
$$u\cdot v=v\cdot u=1/4\cdot(1+j)(1-j)=1/4\cdot(1^2-j^2)=0\,,$$
and that every split-complex number $a+jb$ can be expressed as $c\,u+d\,v$ (with $c=a+b$ and $d=a-b$). Now we have
$$(c\,u+d\,v)^2=c^2\,u+d^2\,v$$
so, in order that A) be satisfied ($(cu+dv)^2=k\,(cu+dv)$), we need $c^2=k\cdot c$ and $d^2=k\cdot d$. That is, either $c=0$ or $c=k$, and the same for $d$: either $d=0$ or $d=k$. It gives all in all $4$ such numbers ($0,k,ku,kv$), but $0$ should be excluded if $z>1$ because of the hypothesis. So, $z$ can be at most $3$, and with $z=3$, $\ S:=\{k,\,ku,\,kv\}$ satisfies the condition: $k\cdot ku=k^2u\notin S$, $\ ku\cdot kv=0\notin S$...
