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!
 A: Here's a fairly clean solution to find one of the cycles:
First simplify the algebra dramatically by substituting $y = -\frac{2i}{101} (z + \frac{101}{2})$, so we can solve for the fixed points of

$$g(y)=\frac{y^2 - 1}{2y}$$

instead (Without the factor of $-i$, you get a plus sign on the numerator - I worked it out with the $-i$, so sticking with that for now. This also gives you factors of $i$ right at the end, which are avoided this way.)
Now, we can compute
$$\begin{align}h(y) := (g \circ g)(y) &= {\left({y^2-1\over2y}\right)^2 - 1\over2\left({y^2-1\over2y}\right)} \\
&=\frac{(y^2 - 1)^2 - 4y^2}{4(y^2-1) y}
\\&=\frac{y^4 - 6y^2 + 1}{4y^3 - 4y}
\end{align}$$
noting that this is an odd function. Observe if $h(y) = -y$, then we have a fixed point of $h \circ h$. But pleasantly, $h(y)$ is odd, so solutions to $h(y) = -y$ should come in $\pm$ pairs, and since we have polynomials this indicates that they should be easy to solve.
$$\begin{align}h(y) &= -y \\
y^4 - 6y^2 + 1 &= -4y^4 + 4y^2 \\
5y^4 - 10y^2 + 1 &= 0\end{align}$$
But the roots of this are easy to find: $5y^4 - 10y^2 + 1 = 5(y^4 - 2y^2) + 1=5(y^2-1)^2-4$, so $y = \pm \sqrt{1 \pm \frac{2}{\sqrt{5}}}$. Now this is certainly a period $2$ point of $h$ since it is non-zero, so gives a genuine period $4$ point of $g$!
Transform back to give 
$$z = -\frac{101}{2} \pm \frac{101i}{2} \sqrt{1 \pm \frac{2}{\sqrt{5}}}$$
a 4-cycle.

Kudos to OP for posting this without the dynamical systems context - that certainly would've put me off this interesting problem!
A: 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 $4$. 
A: Here is my solution, as promised. First, I restate the problem:

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
  $$f(z_1)=z_2$$
  $$f(z_2)=z_3$$
  $$f(z_3)=z_4$$
  $$f(z_4)=z_1$$

Notice that $f$ is equal to
$$f(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2}-\frac{1}{101}}$$
from which it easily follows that
$$f^{\circ n}(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2^n}-\frac{1}{101}}$$
where $f^{\circ n}$ represents $f$ composed $n$ times. In order for $f$ to have a $4$-cycle, $f^{\circ n}(z_1)$ must be a periodic function of $n$ with period $4$. This occurs if and only if
$$\bigg(\frac{1}{z_1}+\frac{1}{101}\bigg)^{2^n}$$
is also a periodic function of $n$. Now recall some properties of the complex roots of unity. If $\omega_k$ is defined as
$$\omega_k=\cos\frac{2\pi}{k}+i\sin\frac{2\pi}{k}$$
then an elementary property is that
$$\omega_k^p=\omega_k^{p\bmod k}$$
Now notice that $2^n\bmod 5$ is periodic with period $4$, with
$$2^{4n}\bmod 5 =1$$
$$2^{4n+1}\bmod 5 =2$$
$$2^{4n+2}\bmod 5 =4$$
$$2^{4n+3}\bmod 5 =3$$
Thus, $f^{\circ n}(z_1)$ should be periodic if we let
$$\frac{1}{z_1}+\frac{1}{101}=\omega_5$$
so that
$$\frac{1}{z_2}+\frac{1}{101}=\omega_5^2$$
$$\frac{1}{z_3}+\frac{1}{101}=\omega_5^4$$
$$\frac{1}{z_4}+\frac{1}{101}=\omega_5^3$$
giving us the values
$$z_1=\frac{101}{101\omega_5-1}$$
$$z_2=\frac{101}{101\omega_5^2-1}$$
$$z_3=\frac{101}{101\omega_5^4-1}$$
$$z_4=\frac{101}{101\omega_5^3-1}$$
...which can be simplified, of course, but you guys have already done that part for me. 
