Every time a real solution. I have got an interesting exercise. Proof that for all positive integer $a$ and $p(x) = x^2+2013x + 1$, $\underbrace{p(p(\dots p}_{a \ \ \text{times}}(x)\dots )) = 0$ has got at least 1 real solution $x_0$.
Hope you can help me. I haven't got any ideas to proof that. Thanks.  
 A: Consider the solutions of 
$p(x)=x$ 
this gives, 
$$x^2+2012x+1=0 \implies x = \frac{-2012-\sqrt{2012^2-4}}{2} , \frac{-2012+\sqrt{2012^2-4}}{2} $$
Let $$ m= \frac{-2012-\sqrt{2012^2-4}}{2} , n=  \frac{-2012+\sqrt{2012^2-4}}{2}$$
Note that, $m<n<0$.
Now let $f(x)= {}\underbrace{p(p(\ldots(p}_{a\ {\rm times}}(x))\ldots))$.
Then, note that, $f(n)=n<0$
And, also the biggest term(term with largest degree) will be $x^{2^a}$ with positive coeffecient. So when $x \rightarrow \infty, f(x) >0$
Since $f(x)$ is a polynomial , so it must be continuous . 
And so there will be a solution of $f(x)=0$ in range $(m,\infty)$ $\Box$
A: Possibly you could show there is $x_1 < 0$ such that $p(x_1)=x_1$. And there is $x_2 > x_1$ such that $p(x_2)>0$. Then for each natural $a$:
$$p^{a}(x_1) = x_1 <0$$
and
$$p^{a}(x_2)>0$$ 
A: $p(x)$ attains its minimum at $(x_{\min}, y_{\min})$, where $x_{\min} = -\frac{2013}{2}$ and $y_{\min} = p(x_{\min}) = x_{\min}^2 + 2013x_{\min} + 1$. The exact values of $x_{\min}$ and $y_{\min}$ are not important; what is important is that $y_{\min} < x_{\min} < 0$, which you can check for yourself.
Now let $q(x)$ be the polynomial $p(x)$ restricted to the domain $[x_{\min},\infty)$. So $q(x)$ is defined on the interval $[x_{\min},\infty)$, and takes the value $p(x)$ on that interval. Because $q(x)$ is monotonic increasing, it has a well-defined inverse on its range, which is $[y_{\min},\infty)$; and because $x_{\min} > y_{\min}$, it has a fortiori a well-defined inverse $q^{-1}(x)$ on $[x_{\min},\infty)$.
So to solve $\underbrace{p(p(\dots p}_{a-times}(x)\dots )) = 0$, just iterate this inverse $a$ times starting from $0$:
$$x = \underbrace{q^{-1}(q^{-1}(\dots q^{-1}}_{a-times}(0)\dots ))$$
