Manipulation of generating functions+quadratic equation with generating functions 
Generating functions really give me a hard time. I'm trying to understand this proof. There are two things I don't see. How do you get the quadratic equation with the $P(z)$ (with straightforward manipulations)? Secondly why can one derive $P(z) = z^2 C(z)^2$?
 A: OP  did not  identify source,  hence  some guesswork  required as  to
context  and objective.  I  get for  plane trees  by  path length  the
following functional equation:
$$T(z,u) = z + z T(uz,u) + z T(uz, u)^2 + \cdots
= z \frac{1}{1-T(uz, u)}.$$
This is because  in the recursive step the subtrees  being attached to
the new  root have their  path lengths  for every node  incremented by
one, plus the root, with zero path length. Looking this up in the OEIS
we  find OEIS  A138158  where we  discover
that there is no simple closed form for $[u^k] [z^n] T(z, u).$ 
We have
$$\sum_{n\ge 1} E_n z^n =
\frac{1}{2} (T(z,1) + T(z,-1))
\quad\text{and}\quad
\sum_{n\ge 1} O_n z^n =
\frac{1}{2} (T(z,1) - T(z,-1))$$
and hence
$$\sum_{n\ge 1} (E_n-O_n) z^n = T(z, -1).$$
The functional equation tells us that
$$T(z, -1) = z \frac{1}{1-T(-z,-1)}
\quad\text{and}\quad
T(-z, -1) = - z \frac{1}{1-T(z, -1)}.$$
Substituting the latter into the former we find
$$T(z, -1) = z \frac{1}{1+z/(1-T(z, -1))}$$
which is
$$T(z, -1) (1 + z - T(z, -1)) = z (1 - T(z, -1))$$
or
$$T(z, -1) (1 + 2z - T(z, -1)) = z.$$
Solving the quadratic yields
$$T(z, -1) = z + \frac{1\pm\sqrt{1+4z^2}}{2}.$$
Here we must choose the second branch so as not to have a coefficient
on $[z^0]$, which contradicts the functional equation. Note also that
with
$$C(z) = \sum_{n\ge 0} C_n z^n = \frac{1-\sqrt{1-4z}}{2z}$$
we have
$$- C(-z) = \sum_{n\ge 0} (-1)^{n+1} C_n  z^n 
= \frac{1-\sqrt{1+4z}}{2z}.$$
which says that
$$T(z, -1) = z - z^2 C(-z^2).$$
Extracting coefficients we get
$$[[n=1]] - [[n\ge 2]] \times [z^{n-2}] C(-z^2)
\\ = [[n=1]] - [[n\ge 2,n\;\text{even}]]\times [z^{n/2-1}] C(-z)
\\ = [[n=1]] + [[n\ge 2,n\;\text{even}]]\times (-1)^{n/2} C_{n/2-1}.$$
This may be verified using the Lagrange-Burmann  formula with
$$z = \frac{T(z, -1)\times (1-T(z,-1))}{1-2T(z,-1)}$$
so that we get
$$[z^n] T(z, -1) =
\frac{1}{n} [w^{n-1}] \frac{(1-2w)^n}{(1-w)^n}.$$
This is
$$\frac{1}{n} \sum_{q=0}^{n-1} 
{n\choose q} (-1)^q 2^q {2n-2-q\choose n-1}.$$
We get for the sum
$$\frac{1}{n}
\sum_{q=0}^{n-1} {n\choose q} (-1)^q 2^q
[v^{n-1-q}] (1+v)^{2n-2-q}
\\ = \frac{1}{n} [v^{n-1}] (1+v)^{2n-2}
\sum_{q=0}^{n-1} {n\choose q} (-1)^q 2^q
 v^q (1+v)^{-q}.$$
Here we get zero contribution  to the coefficient extractor when $q\gt
n-1$ and hence  we may extend the sum to  infinity (actually extension
to $n$ is sufficient):
$$\frac{1}{n} [v^{n-1}] (1+v)^{2n-2}
\left(1-\frac{2v}{1+v}\right)^{n}
\\ = \frac{1}{n} [v^{n-1}] (1+v)^{n-2} (1-v)^{n}
= \frac{1}{n} [v^{n-1}] (1-v^2)^{n-2} (1-2v+v^2).$$
Extracting  coefficients from  this we  get for  $n=1$ the  value one,
which is correct.  For $n\ge 2$ with $n$ odd we obtain
$$\frac{1}{n} (-1)^{(n-1)/2} {n-2\choose (n-1)/2}
+ \frac{1}{n} (-1)^{(n-3)/2} {n-2\choose (n-3)/2} = 0.$$
For $n$ even we obtain
$$-\frac{2}{n} (-1)^{(n-2)/2} {n-2\choose (n-2)/2}
= (-1)^{n/2} \frac{1}{(n-2)/2+1} {n-2\choose (n-2)/2} 
\\ = (-1)^{n/2} C_{n/2-1}.$$
This is the claim. 
The Maple code  that was used to  verify the above and  query the OEIS
goes as follows.

with(combinat);

X :=
proc(n)
option remember;
local part, psize, mset, res;

    if n=1 then return 1 fi;

    res := 0;

    part := firstpart(n-1);

    while type(part, `list`) do
        psize := nops(part);
        mset := convert(part, `multiset`);

        res := res + u^(n-1)*
        mul(X(v), v in part)*
        psize!/mul(v[2]!, v in mset);

        part := nextpart(part);
    od;

    expand(res);
end;

EODIFF := n -> subs(u = -1 , X(n));

CAT := n -> 1/(n+1)*binomial(2*n,n);

EODIFFX := n ->
`if`(type(n,odd), `if`(n=1, 1, 0), (-1)^(n/2)*CAT(n/2-1));

