$P(n)-2015 \equiv -2015 ( mod P(n) )$ so $P(P(n)-2015) \equiv P(-2015) ( mod P(n) )$ but $P(P(n)-2015) \equiv 0 ( mod P(n) )$ therefore $ P(-2015) \equiv 0 ( mod P(n) )$ which means that for every $n$ , $P(n)$ divides $P(-2015)$ which means (if $P(-2015)$ differs from $0$) that $P(n)$ is constant. Contradiction therefore $P(-2015)=0$

Notice this: $P(P(n) + n)$ is divisible by $P(n)$. So we have $P(P(n) - 2015) - P(-2015)$ is divisible by $P(n)$ Hence, $P(-2015)$ is divisible by $P(n)$. On other hand, $P(x)$ isn't constant. Setting $P(x) = a_nx^n + ... + a_0 (a_n, n > 0)$ $$\lim_{x\to\infty}P(x) = \infty$$Take $n$ large enough then $P(-2015) = 0$. P.s: I dont know why do we need 'positive ineger coefficients'

Just use the fact P(n+d)$\equiv$P(n)(mod d) $\forall$ n,d $\in$Z if P(x) is a non constant polynomial with integer coefficients and we are done.

Got a staright forward method. let P(n)-2015=K P(n)=K+2015 A/C to question let a*P(n)=P(P(n)-2015) a*(K+2015)=p(k) for k= -2015 p(-2015)=0

Let $P(x) = a_nx^n+a_{n-1}x^{n-1}+\dots a_1x+a_0$. So put $x = P(n) - 2015$, to get: \[ a_n(P(n)-2015)^n+a_{n-1}(P(n)-2015)^{n-1}+\dots+a_1(P(n)-2015)+a_0 \]Now taking $\pmod{P(n)}$, we get: \[ a_n(-2015)^n+a_{n-1}(-2015)^{n-1}+\dots+a_1(-2015)+a_0 = P(-2015) \equiv 0 \pmod{P(n)} \quad \forall n \in \mathbb{N} \]This concludes the proof.

Alright, its time to post an overkill solution. But, it is what it is... Let me thank this class in OMC conducted by Nuterrow. Solution: Here's the main lemma which will kill the problem on the spot. Lemma: If two polynomials $f,g \in \mathbb{Z}[x]$ are such $f(n) \mid g(n)$ for infinitely many integers $n$, then $f(x) \mid g(x)$(polynomial wise). Proof: Due to size reasons, it is easy to see $\deg(g) > \deg(f)$. By the division algorithm over $\mathbb{Q}[x]$, write $g = fq + r$ for $f,q \in \mathbb{Q}[x]$ and $\deg(r) < \deg(f)$. One can clear out the denominators by multiplying by suitable integer, so we assume that all the polynomials are in $\mathbb{Z}[x]$. Just pick huge $k$ such that $f(k) \mid g(k)$. It would actually mean that $f(k) \mid r(k)$. But since $\deg(r) < \deg(f)$, we must have $r \equiv 0$ as desired. $\square$ In our case, applying the theorem here we will get that there exists some polynomial $Q \in \mathbb{Q}[x]$ such that $P(x)Q(x) = P(P(x)-2015)$ for all $x \in \mathbb{C}$. By Fundamental Theorem of Algebra, pick a root $\alpha \in \mathbb{C}$ of $P$. This will immediately $P(-2015) = 0$ as desired. $\blacksquare$