Rewinding 2016 Mathematically

Here is my solution.I devide it into three parts so as to present the whole idea more clearly.

Part 1. The real roots of $f(x)$ are $x_{1}=\frac{-1-\sqrt{3}}{2}$ , $x_{2}=\frac{-1+\sqrt{3}}{2}$ .

So $f(f(x)) =0\Leftrightarrow$ $f(x)=x_{1}$ or $f(x)=x_{2}$

$\Leftrightarrow$ $2x^2+2x-1-x_{1}=0$ or $2x^2+2x-1-x_{2}=0$

(now,note that $x_1+x_2=-1$ )

$\Leftrightarrow$ $2x^2+2x+x_{2}=0$ or $2x^2+2x+x_{1}=0$

Since $x_{1,2}<\frac{1}{2}$ ,the discriminant $△=4-8x_{1,2}>0$ ,which means the equations above both have two distinct real roots.

Assume that they have a common root,say, $c$ ,then we derive $2c^2+2c+x_{2}=0$ and $2c^2+2c+x_{1}=0$ and so $x_{1}=x_{2}$ ,which is a contradiction.Therefore,they cannot have common roots.

Part 2. You might have noticed that the result above can be generalized.

Let's prove(by induction) that

$f^{(n)}(x)$ has $2^{n}$ distinct real roots $x_1,x_2,\cdots,x_{2^n}$ satisfying : for $i=1,2,\cdots,2^n,$

$(1) x_i\in ( -\frac{3}{2}, \frac{1}{2})$ ,

$(2) x_i+x_{2^n+1-i}=-1$ .

The case $n=1$ is obvious. Now assume that the case $n=k$ is true. For $n=k+1$ , $f^{k+1}(x)=0$ $\Leftrightarrow$ $f^{k}(f(x))=0$ $\Leftrightarrow$ $f(x)=x_i, i=1,2,\cdots,2^k$ (then we'll get $2^k$ sets of roots,and the union set of them is formed exactly by all the roots of $f^{k+1}(x)$ ) $\Leftrightarrow$ $2x^2+2x-1-x_{2^k+1-i}=0,i=1,2,\cdots,2^k$ $\Leftrightarrow$ $2x^2+2x+x_{i}=0,i=1,2,\cdots,2^k$ .

Since $x_i<\frac{1}{2} ,\forall i$ ,the discriminant $△=4-8x_{i}>0$ ,which means each quadratic equation has two distinct real roots(their sum is $-\frac{2}{2}=-1$ ).And it's clear that no two equations share a common root(see part1,it's similiar).

Besides, $\forall i$ ,the roots of $2x^2+2x+x_{i}=0$ satisfy

$-\frac{3}{2}=-\frac{1}{2}-\frac{1}{2} \sqrt{1-2(-\frac{3}{2})}<-\frac{1}{2}-\frac{1}{2} \sqrt{1-2x_i}<-\frac{1}{2}+\frac{1}{2} \sqrt{1-2x_i}<-\frac{1}{2}+\frac{1}{2} \sqrt{1-2(-\frac{3}{2})}=\frac{1}{2}$

Hence,the case $n=k+1$ is also true.

Part 3. Since $f(0)=-1,f(-1)=-1,f^n(-1)=-1(\forall n),$
0 cannot be root of any $f^n(x)$ . Denote the number of negative real roots of $f^{n}(x)$ by $a_n$ .Denote the number of positive real roots of $f^{n}(x)$ by $b_n$ .

Consider the equation $2x^2+2x+x_{i}=0$ .Since it has been proven to has two distinct real roots,whose sum and product is $-1$ and $x_i$ respectively,

if $x_i<0$ ,then it has one negative real root and one positive real root; if $x_i>0$ ,then it has two negative real roots.

Thus $a_{n+1}=a_n+2b_n$ , $b_{n+1}=a_n$ ,for every positive integer $n$ .From this result we derive $a_1=1$ , $a_2=3$ and $a_{n+2}=a_{n+1}+2a_n$ ,for every positive integer $n$ . The characteristic equation of the sequence is $x^2-x-2=0$ ,whose roots are $2$ and $(-1)$ . Hence $a_n=\frac{2^{n+1}+(-1)^n}{3}$ for all n. So $a_{2016}=\frac{2^{2017}+1}{3}$ .

The answer is $2+2017+3=\boxed{2022}$