If $P(n)$ divides $P(P(n)-2015)$, prove that $P(-2015)=0$ 
Q. Let $P(x)$ be a non-constant polynomial whose coefficients are
  positive integers. If $P(n)$ divides $P(P(n)-2015)$ for every natural
  number $n$, prove that $P(-2015)=0$.

In one of the sources, the solution given is as follows:
Note that $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$ for every positive integer $n$. But $P(n)$ divides $P(P(n)-2015)$ for every positive integer n. 
Therefore, $P(n)$ divides $P(-2015)$ for every positive integer $n$. Hence $P(-2015)=0$.
I am not able to understand that how $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$.
Please help me out.
 A: $\bmod \color{#c00}{P(n)}\!:\ 0\equiv P(\color{#c00}{\overbrace{P(n)}^{\large \equiv\ 0}}-2015)\equiv P(-2015)\ $ by the Polynomial Congruence Rule. 
Thus $\, P(n)\mid P(-2015)\, $ for all $n$ so $\,P(-2015) = 0\,$ since $P$ is nonconstant so unbounded. 
Remark $ $ Without congruences, put $\,x = P(n),\, a = -2015\,$ below
$\quad$ if $\,x\,$ divides $\,P(x\!+\!a)\,$ then $\,\underbrace{x\ {\rm divides}\  P(x\!+\!a)-P(a)}_{\rm\large Factor\ Theorem}\ $ so $x$ divides $P(a) = $ their difference
A: Note that $P(n)=P(n)-2015+2015$
$P(n)=P(n)-2015-(-2015)$ which divides $P(P(n)-2015)-P(-2015)$ for all positive integers $n$
Note that $P(n)$ divides $P(P(n)-2015)$ for all positive integers $n$
Finally, we can say that $P(n)$ divides $P(-2015)$ for all positive integers $n$. 
So, $P(-2015)=0$
A: Let $b$ be an arbitrary integer of $\mathbb Z$. The notations below are clear to understand.
$$P(x)=\Sigma_{i=0}^{i=d}a_ix^i\\P(n)=\Sigma_{i=0}^{i=d}a_in^i\\P(P(n)-b)=\Sigma_{i=0}^{i=d}a_i(P(n)-b)^i\\P(P(n)-b)=P(n)Q(n)+\Sigma_{i=0}^{i=d}(-1)^ia_ib^i\\P(P(n)-b)=P(n)Q(n)+P(-b)$$ Since the constant $P(-b)=c$ must be divisible by $P(n)$ for all $n$ and $|P(n)|$ is unbounded, we must have $P(-b)=0$. The property is true for all constant $b$, in particular for $b=2015$. 
