Polynomials and more polynomials?

Since the value for this expression: $(p-1)^{2} - Q_{2014}(p)=\left(p+1\right)^2-p$ is entirely dependent on $p$ , we can assume the least value for $p$ to get the answer.

Its a quadratic whose minimum is when $x=-0.5$ so it will be ever increasing if $x$ is positive. So we can make $p=1$ and prove that it works.

I am going to sketch a prove that if $Q(x)$ is strictly increasing or decreasing from $x>p$ onward, and that one of the root for $Q(x)-x=0$ is $x=p$ , where $p$ is a positive integer, one of the root for $Q_{n}(x) - x = 0$ would be $x=p$ , where $n$ is a positive integer more that $1$ .

When $Q(x)$ is strictly increasing from $x>p$ onward, when $Q(x)>x$ || $(x>p)$ , $Q_{n+1}(x)>Q_{n}(x)$

When $Q(x)<x$ || $(x<p)$ , $Q_{n+1}(x)<Q_{n}(x)$

So the equilibrium is reached when $Q(x)=x$ , $Q_{2015}(x)=Q(x)=x$

Note that the degree for the polynomial $Q(x)$ has to be more than $1$ in order for this to work.

A similar thing can be done for the decreasing part.

A strictly increasing function $Q(x)$ from $x>1$ onward such that $Q(x)-x=0$ has roots $1$ can be $Q(x)=x^{2}$ , there are infinitely many solutions such as $Q(x)=x^2-2x+2$ . Anyways, trying out $P_{n}(x)-x=0$ , where $P(n)=x^{2}$ it indeed gives the root $1$