Yet another Question on Number Theory Let $P(x)$ be a non-zero polynomial with integer coefficients. If $P(n)$ is divisible by $n$ for each positive integer $n$, what is the value of $P(0)$?
EDIT: The answer is coming out to be zero with an example I know it is obvious but is there any mathematical proof for this?
 A: Let us show that the assumption implies that $x\,|\,P(x)$.
In general, we must have $$P(x)=x\;Q(x)+c$$
Where $Q(x)$ is another poynomial with integer coefficients (the quotient) and $c$ is an integer constant, the remainder.  
Now we remark that  $$n\,|\,P(n)\implies n\,|\, c$$
But if $c$ were non-zero this could only be true for finitely many $n$.  As the assumption is that it is true for all positive $n$ then $c$ must be $0$.  Thus $P(x)=x\;Q(x)$ so $P(0)=0$.
A: Since $a-b|P(a)-P(b)$ when $a,b \in \mathbb{Z}$ and $P(x) \in \mathbb{Z[x]}$
we have $(n-0)|P(n)-P(0) \Rightarrow n|P(0) \ \forall n \in \mathbb{N}$
For sufficiently large $n$ we immediately get $P(0) = 0$
A: $P(0)=0$ because if
$P(x)=a_n x^n+a_{n-1} x^{n-1}+\ldots+a_2 x^2+a_1 x+a_0$
$P(n)=a_n n^n+a_{n-1} n^{n-1}+\ldots+a_2 n^2+a_1 n+a_0$ is  a multiple of $n$ for any integer $n$ only if $a_0=0$
$P(n)=n\left(a_n n^{n-1}+a_{n-1} n^{n-2}+\ldots+a_2 n+a_1 \right)$
A: $P(x)=xQ(x)+P(0)$.
Note that for any prime number $p$, $pQ(p)\equiv 0(\mod p)$.
Then $$P(p)-P(0)\equiv 0(\mod p)\Rightarrow P(p)\equiv P(0)(\mod p)\\\Rightarrow P(0)\equiv 0(\mod p)$$
This is true for all $p$. Hence $P(0)=0\space\space\space\space\blacksquare$
