System of quadratic equations with parameter

Question

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. :(

Answer 1

• 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 $$a^2+a-p=0$$

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

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• 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

$$a^2-a+1-p=0$$

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

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

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• 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 $4$.