Let $P(x)=x^2 +2013x+1$. Show that $P(P(\cdots P(x))\cdots)=0$ (i.e. $P$ is $n$-times nested) has at least one real root for any $n$ $P$.
For $n = 1$ this is obvious. Next, for $n = 2$ we get a fourth order polynomial $$x^4 + 4,026 x^3 + 4,054,184 x + 2,015$$ and by substituiting $y = x + 2,013/2$ can elimate the cubic term. Rearranging and standardizing yields a 4th order equation $$y^4 – 4,048,139/2 y^2 + 16,387,413,154,661/16 = 0$$ It can be solved but further substitutions quickly become intractable and don´t lead anywhere. Can anyone help? Thanks.