Quadratic Functional equations. Suppose $f$ is a quadratic ploynomial, with leading cofficient $1$, such that $$f(f(x) +x) = f(x)(x^2+786x+439)$$ For all real number $x$. What is the value of $f(3)$?
 A: Suppose $f(x) = x^2 + bx + c$
Then 
$$f(f(x)+x) = (f(x)+x)^2 + b(f(x)+x) + c = f(x)^2 + 2xf(x) + x^2 + bf(x) + bx + c = f(x)^2 + 2xf(x) + bf(x) + f(x) = f(x)(f(x) + 2x + b+ 1)$$
Equating factors yields
$$f(x) + 2x + b+ 1 = x^2 + 786 x + 439$$
Then letting $x=3$ we have
$f(3) = 2799-b$
Finding $b$ is a trivial matter by expanding out the previous equation.
There may be an indirect method for determining $b$ but this is as far as I can get without solving for $f(x)$.
A: Stupid method: 
Let $f(x) = x^2 + ax + b$. Then the left hand side of the equation is 
$$\begin{split}
f(f(x) + x) & =f( x^2 + ax + b + x) \\
&= (x^2+ (a+1)x +b)^2 + a(x^2+(a+1) x+ b) + b
\end{split}$$
If you compare the $x^3$ coefficient, this gives $2(a+1) = 786 +a$. Comparing the constant coefficient tells you $b$. 
A: Brute forse. Let $f(x)=x^2+a x+b.$ Then 
$$
f(f(x)+x)-f(x)({x}^{2}+786\,x+439)= \\=\left( -784+a \right) {x}^{3}+ \left( b-438-783\,a+{a}^{2} \right) {x
}^{2}+ \left( {a}^{2}-784\,b+2\,ab-438\,a \right) x-438\,b+{b}^{2}+ab=0.
$$
By solving the corresponding system we get $a=784, b=-346.$  Then $f(3)=2015.$
