<s>[INST] Recreational math problem: $f(x)=\frac{x^2}{2x+101}$ Here is a math problem (just for fun) for the residents of MSE to enjoy:

Let $f(z)$ be defined as 
  $$f(z)=\frac{z^2}{2z+101}$$
  Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that
  $$f(z_1)=z_2$$
  $$f(z_2)=z_3$$
  $$f(z_3)=z_4$$
  $$f(z_4)=z_1$$

The answer (my answer, at least) is a bit messy, so watch out! After a correct answer is posted, I will post my solution method.
Enjoy!
 [/INST] HINT:
Let's do some reductions of $f$: 
$$f(101 x) = \frac{(101 x)^2}{2\cdot 101 x + 101}= 101\cdot \frac{x^2}{2x+1}$$
so we have
$$f= s \circ g \circ s^{-1}$$ where $s(x) = 101 x$ and $g(x)=\frac{x^2}{2x+1}$. Further
$$g(x)=\frac{x^2}{2x+1}= \frac{x^2}{(x+1)^2 - x^2}= \frac{1}{(1+\frac{1}{x})^2 -1}= t^{-1}\circ h \circ t(x)$$
where $t(x) = \frac{1}{x}+1$ and $h(x)=x^2$. Finally 
$$f=s \circ t^{-1} \circ h \circ t \circ s^{-1}= u \circ h \circ u^{-1}$$
where $u(x)=s\circ t^{-1}(x)=\frac{101}{x-1}$. In other words, $f\circ u = u\circ h$,
$$f(\frac{101}{x-1})=\frac{101}{x^2-1}$$
So $f(x)$ is conjugate to the function $h(x)=x^2$. One checks easily that $x$ starts a cycle of length $4$ for $h$ if and only if $u(x)$ starts a cycle of length $4$ for $f$. 
To find cycles of length $4$ for $h$, one  looks for solutions of 
$x^{16}=x$ that are not solutions of $x^4=x$. That means, $x$ satisfies $x^{15}=1$ but not $x^{3}=1$. There are $15-3=12$ such solutions, 
$$x= \exp \frac{2 k \pi i}{15}, \ \ k = 1,2,3,4,6,7,8,9,11,12,13,14$$
They break into $3$ cycles of length $4$. 
</s>