Finding functional equation for generating function I'm given
$$
a_n = \sum_{i=2}^{n-2} a_ia_{n-i} \quad (n\geq 3), a_0 = a_1=a_2 = 1
$$
and I need to find the functional equation for the generating function satisfying the above equality. I obtained the correct answer, but I don't understand why some of the steps are allowed.
$$
\begin{align}
f(x) &= \sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2} a_i a_{n-i} x^n\\
&=\sum_{n=0}^{\infty} \sum_{i=2}^{n-2}a_ix^i a_{n-i}x^{n-i}\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{n-2} a_ix^i a_{n-i}x^{n-i} \text{ (since it doesn't make sense for $n$ to start at $0$})\tag1\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n-i}x^{n-i} \text{ (since terms past $n-2$ don't exist)} \tag2\\
&=\sum_{n=2}^{\infty} \sum_{i=2}^{\infty} a_ix^i a_{n}x^{n} \text{ (not sure)} \tag3
\end{align}
$$ 
Therefore,
$$
f(x) -1-x-x^2 = (f(x) -1 -x)^2.
$$
So the equations $(1),(2)$, and $(3)$, I'm not sure why those steps are allowed (or if they're even correct). Could someone please help? 
 A: For this problem, I would go in the other direction:
$\begin{align}
f^2(x) 
&= (\sum_{n=0}^{\infty} a_n x^n) (\sum_{m=0}^{\infty} a_m x^m)\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n x^n  a_m x^m\\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n a_m x^{n+m}\\
&= \sum_{i=0}^{\infty} \sum_{j=0}^{i}a_j a_{i-j} x^{i}\\
&= \sum_{i=0}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i \sum_{j=0}^{i}a_j a_{i-j} \\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +\sum_{j=2}^{i-2}a_j a_{i-j} )\\
&= a_0^2+ 2 a_0 a_1 x + \sum_{i=2}^{\infty} x^i (2a_0 a_i + 2 a_1 a_{i-1} +a_i )\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1\sum_{i=2}^{\infty} x^i  a_{i-1}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)\sum_{i=2}^{\infty} x^i a_i + 2 a_1 x\sum_{i=2}^{\infty} x^{i-1}  a_{i-1}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x\sum_{i=1}^{\infty} x^{i}  a_{i}\\
&= a_0^2+ 2 a_0 a_1 x + (2 a_0+1)(f(x)-a_0-x a_1)+ 2 a_1 x (f(x)-a_0)\\
&= a_0^2+ 2 a_0 a_1 x + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)- 2 a_0 a_1 x\\
&= a_0^2 + f(x)(2 a_1 x+2 a_0+1)-(2 a_0+1)(a_0x+ a_1)\\
&= 1 + f(x)(2 x+3)-3(x+1)\\
&= f(x)(2 x+3)-3x-2\\
\end{align}
$
This gives a quadratic equation for $f(x)$
(as expected, from the recursion for $a_n$):
$f^2 - (2x+3)f + 3x+2 = 0$.
Solving,
the discriminant is
$(2x+3)^2 - 4(3x+2)
= 4x^2+12x+9-12x-8
=4x^2+1
$,
so the solutions are
$f(x) = \frac{2x+3 \pm \sqrt{4x^2+1}}{2}
$.
Since $f(0) = 1$, 
we have to take the minus sign,
so
$f(x) = \frac{2x+3 - \sqrt{4x^2+1}}{2}
$.
There is a significant chance that
there is some error here 
(since I did it off the top of my head),
but the basic idea is that
this type of recurrence  always leads to
a quadratic equation in the generating function.
Note that the coefficients of all the odd powers of $x$
(from $x^3$ on)
are zero, which agrees with Glenn O's comment
about $a_3$.
Also, I do not know whether or not it is a coincidence
that the discriminant does not have an $x$ term - 
if it did, expanding the square root would be a lot harder.
