System of quadratic equations with parameter I'd appreciate your help with this problem:
$p - a^2 = b$
$p - b^2 = c$
$p - c^2 = d$
$p - d^2 = a$
Where $a$, $b$, $c$, $d$ are real numbers, $p$ is a real parameter lower or equal to 1 and greater or equal to 0. 
Thank you a lot. I am capable of solving the problem for positive numbers, but that's it. :(
 A: *

*Subtracting two adjacent equations,


$$
\left \{
  \begin{array}{ccc}
    (a-b)(a+b) &=& c-b \\
    (b-c)(b+c) &=& d-c \\
    (c-d)(c+d) &=& a-d \\
    (d-a)(d+a) &=& b-a
  \end{array}
\right.$$ 
Note that
$$a=b \iff b=c \iff c=d \iff a=d$$
Now
$$a^2+a-p=0$$

$$a=b=c=d=\frac{-1\pm \sqrt{1+4p}}{2}$$




*

*Subtracting alternate equations, $$
\left \{
  \begin{array}{ccc}
    (a-c)(a+c) &=& d-b \\
    (b-d)(b+d) &=& a-c
  \end{array}
\right.$$ 


Note that
$$a=c \iff b=d$$
$$
\left \{
  \begin{array}{ccc}
    p-a^2 &=& b \\
    p-b^2 &=& a
  \end{array}
\right.$$
Now
$$p-(p-a^2)^2=a$$
$$p^2-(2a^2+1)p+(a^4+a)=0$$
$$(a^2+a-p)(a^2-a+1-p)=0$$
Since $a^2+a-p=0$ reproduces the previous case, we only need to solve
$$a^2-a+1-p=0$$

$$a=c=\frac{1\pm \sqrt{4p-3}}{2}$$
$$b=d=\frac{1\mp \sqrt{4p-3}}{2}$$




*

*For distinct $a$, $b$, $c$ and $d$, that'll be a solution of $12$-th order polynomial equations in terms of $p$.


P.S.
$a,b,c,d$ are critical points of the iteration $u_{n+1}=p-u_{n}^2$ of period $4$.
