MHB Are There Limited Ordered Triples for Complex Number Equations?

[anemone]

Here is this week's POTW:

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Prove that there are only a finite number of possibilities for the ordered triple $(a-b,\,b-c,\,c-a)$ where $a$, $b$ and $c$ are complex numbers satisfying the simultaneous equations

$a(a-1)+2bc=b(b-1)+2ca=c(c-1)+2ab$

and list all such triples.

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[anemone]

Congratulations to castor28 for his correct solution, which you can find below:):

Partial credit goes to the following members for their partially correct solution:
1. Olinguito
2. kaliprasad

Solution from castor28:

Spoiler

The two independent equations can be written as:
\begin{align*}
(a-b)(a+b-2c-1) &= 0 = (a-b)((b-c)-(c-a)-1)\\
(a-c)(a+c-2b-1) &= 0 = -(c-a)((a-b)-(b-c)-1)
\end{align*}
Writing $x=a-b$, $y=b-c$, $z=c-a$, we get:
\begin{align*}
x(y-z-1) &= 0\\
y(z-x-1) &= 0
\end{align*}
Since $x+y+z=0$, we may substitute $z=-x-y$, and we obtain:
\begin{align*}
x(x+2y-1) &= 0\\
y(y+2x+1) &= 0
\end{align*}
We can choose one of the two factors in each equation in four possible ways; in each case, we obtain a non-singular system of two linear equations in $x$ and $y$. Using the fact that $z=-x-y$, we obtain the four possible triples $(x,y,z)$ : $(0,0,0)$, $(0,-1,1)$, $(1,0,-1)$, and $(-1,1,0)$.