Consider this system of $12$ equations $$ \left\{\begin{array}{rcrclr} \alpha^{2}p_{i} & + & \left(1 - \alpha\right)^{2}\left(1 - p_{i}\right) & = & c_{i}, & \forall i =1,2,3,4 \\[1mm] \alpha\left(1 - \alpha\right)p_{i} & + & \left(1 - \alpha\right)\alpha\left(1 - p_{i}\right) & = & d_{i}, & \forall i =1,2,3,4 \\[1mm] \left(1 - \alpha\right)^{2}p_{i} & + & \alpha^{2}\left(1 - p_{i}\right) & = & e_i, & \forall i =1,2,3,4 \end{array}\right. $$ where
$\alpha \in \left[0,1\right]$
$p_{i} \in \left[0,1\right]$ $\forall i = 1, 2, 3, 4$
$c_{i}, d_{i}, e_{i}$ are real numbers $\forall i = 1, 2, 3, 4$.
I want to show that this system of equation has ( or does not have ) a unique solution with respect to $\alpha, p_{1}, p_{2}, p_{3}, p_{4}$. Could you help ?.
This is what I have tried and where I'm stacked. Let $i = 1$. From the second equation, we get $$ \alpha - \alpha^{2} = d_{1} $$ which gives $$ \alpha_{\left(1\right)} = \frac{1 + \sqrt{1 - 4d_{1}}}{2},\quad \alpha_{\left(2\right)} = \frac{1 - \sqrt{1 - 4d_{1}}}{2} $$ From the first equation one can get $p_{1}$. From other equations, I guess one can analogously obtain $p_{2}, p_{3}, p_{4}$.
Is this sufficient to show that the system does not have a unique solution ? Or, is there a way to exclude one between $\alpha_{\left(1\right)},\alpha_{\left(2\right)}$ ?.
As you noticed, second four equations reduce to $\alpha-\alpha^2=d_i$. So the necessary condition for the system to have a solution is $0\le d=d_1=d_2=d_3=d_4\le \frac 14$. The remaining equations reduce to $$p_i(2\alpha-1)=c_i-(\alpha-1)^2=\alpha^2-e_i.$$ It follows $2\alpha^2-2\alpha=c_i+e_i-1=-2d_i$. This is an other necessary condition for the system to have a solution. We assume that both groups of necessarily conditions hold. Now the following cases are possible.
1)) $d=\tfrac 14$. Then $\alpha=\tfrac 12$. Then $p_i$ are undetermined by the system, and it has a solution (not unique) iff $e_i=\alpha^2=\frac 14$ for each $i$
2)) $0\le d<\frac 14$. Then there are two possible choices $\alpha_1$ and $\alpha_2$ for $\alpha$ and
$$p_i=\frac{\alpha^2-e_i}{2\alpha-1}=\frac{\alpha-d-e_i}{2\alpha-1}=\frac 12+\frac{1/2-d-e_i}{2\alpha-1}.$$
We have $p_i\in [0,1]$ iff $|1-2d-2e_i |\le |2\alpha-1|=|2\alpha_j-1|$ for each $i$. If this condition fails for some $i$, then the system has no solutions. Otherwise it has two solutions, one for each $\alpha_j$.