System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$ Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$
I have a solution 
Let $f(x)=\frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$
1) fixed points of $f(x)$
Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
These fixed points are solution of $x^3-x+t=0$ and so :
If $t\in(0,\frac 2{3\sqrt 3})$ : three such solutions
If $t=\frac 2{3\sqrt 3})$ : two such solutions
If $t\in(\frac 2{3\sqrt 3},+\infty)$ : one such solution
2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
These fixed points are solution of $x^2-tx-1=0$ and so :
Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)
3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
These are solutions of a degree $12$ or degree $11$ ugly polynomial :
$(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
$+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
$+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
$+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
$+(t^8-6t^6+11t^4-6t^2+1)$
And I dont know how to determine the number of real roots of this polynomial :
For certain values (for example $t=\frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))
For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.
There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.
But I did not see it up to now. 
 A: I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:
\begin{align*}
12 \quad &\text{when}\; 0<t<\frac{\sqrt{5}-1}{2}\\
11 \quad &\text{when}\; t=\frac{\sqrt{5}-1}{2}\\
12 \quad &\text{when}\; \frac{\sqrt{5}-1}{2}<t<0.6432\\
8 \quad &\text{when}\; t \approx 0.6432\\
4 \quad &\text{when}\; 0.6432<t<\frac{\sqrt{5}+1}{2}\\
3 \quad &\text{when}\; t=\frac{\sqrt{5}+1}{2}\\
4 \quad &\text{when}\; \frac{\sqrt{5}+1}{2}<t<2\\
2 \quad &\text{when}\; t=2\\
0 \quad &\text{when}\; t>2\\
\end{align*}
When $t\approx0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.
When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.
$$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$
There are two vertical asymptotes at $t=\frac{\sqrt{5}\pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):
$$
\frac{y-x}{y}=\left(\frac{x\left(1-\left(\frac{x}{1-y^2}\right)^2\right)^2}{\left(1-\left(\frac{x}{1-y^2}\right)^2\right)^2-x^2}\right)^2
$$
As $y \to\infty$, the left-hand side goes to 1 and $\frac{x}{1-y^2} \to 0$. Therefore:
$$
\pm x\left(1-0^2\right)^2=\left(1-0^2\right)^2-x^2
$$
A: If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.

Assume $1<t$, then you can have infinite negative solutions as per the path
shown.   You cannot have a positive value for any one of the unknowns.
Only if $0<t<<1$ you may find  infinite positive solutions starting with a high value for $a$.
