How to find the closed solution for \begin{cases} a^2+bc=-d\\ d^2+bc =-a\\ b(a+d)=b\\ c(a+d)=c\\ \end{cases}?

Question

Given the following equations.

\begin{cases} a^2+bc=-d\\ d^2+bc =-a\\ b(a+d)=b\\ c(a+d)=c\\ \end{cases}

My attempt:

Eliminating $bc$ in the first two equation, I have \begin{align} a^2-d^2 &= a -d\\ (a-d)(a+d-1)&=0\\ \end{align}

The solution is $a=d$ or $a+d=1$.

The remaining two equations, I have $b=0$ or $c=0$ or $a+d=1$.

Now I am confused how to construct the final answer.

For example, if I chose $a=d=\lambda$ then $bc=-\lambda^2-\lambda$ which is contradictory to either $b=0$ as $a+d=2\lambda\not=1$ or $c=0$ as $a+d=2\lambda\not=1$.

Trying again:

$$ \{(a,b,c,d)|(0,0,0,0),(-1,0,0,-1),(s,t,u,1-s)\} $$ where $s,t,u\in \mathbb{R}$.

Answer 1

If $c$ is nonzero, then we obtain $$ b=\frac{-a^2+a-1}{c},\; d=1-a. $$ If $c=0$, then either $a=b=d=0$, or $b=0$, $a=d=-1$, or $b$ is arbitrary with either $$ a=\frac{1-\sqrt{-3}}{2}, d=\frac{1+\sqrt{-3}}{2} $$ or $a$ and $d$ interchanged.

Remark: This seems to be related to this post.

Answer 2

From you equations, if $a=d$, then $bc=0$ means either $a=d=0$ or $a=d=-1$. The other solution is $a=d=\frac12$, so $bc=-\frac34$. For $a+d=1$ you have $$a^2+1-a+bc=0$$ and similar for $d$, where $b$ and $c$ can take any values. $$a_{1,2}=\frac{1\pm\sqrt{1-4(1+bc)}}{2}$$ So you have $d_{1,2}=a_{2,1}$.