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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?

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  • $\begingroup$ what do you think it is, and why? $\endgroup$ – uniquesolution Aug 8 '17 at 13:28
  • $\begingroup$ Have you tried anything? Can you, for example, come up with some examples of such a $P(x)$? $\endgroup$ – lulu Aug 8 '17 at 13:29
  • $\begingroup$ @lulu With an example the answer is coming out to be zero but i was wondering if there was any mathematical proof $\endgroup$ – Dhruva Aug 8 '17 at 13:33
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    $\begingroup$ Well, write out some examples for $P(x)$. Notice a pattern? As a general way to get started, it is much easier to prove or disprove a specific statement than it is to investigate an unknown situation. If you believe $P(0)=0$, good. Try to prove that. That is the same as saying that $x$ divides $P(x)$. Well...suppose $x$ did not divide $P(x)$. What would that tell you? $\endgroup$ – lulu Aug 8 '17 at 13:34
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    $\begingroup$ There is indeed a very easy proof. Write down an arbitrary polynomial $P$ and compute $P(0)$ and $P(n)$ in terms of the coefficients. Then investigate the condition $n\mid P(n)$. $\endgroup$ – Dietrich Burde Aug 8 '17 at 13:35
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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 $P(0)=0$.

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  • $\begingroup$ Could you please cancel the downvote. I have corrected my proof :) $\endgroup$ – Raffaele Aug 8 '17 at 13:59
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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 $P(0) = 0$

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  • $\begingroup$ why the first line is true??? what if $a<b$ $\endgroup$ – MAN-MADE Aug 8 '17 at 14:17
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$P(0)=0$ 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 $a_0=0$

$P(n)=n\left(a_n n^{n-1}+a_{n-1} n^{n-2}+\ldots+a_2 n+a_1 \right)$

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    $\begingroup$ You just pick an example, how is this even accepted!!! $\endgroup$ – MAN-MADE Aug 8 '17 at 13:43
  • $\begingroup$ This is only an example, not a proof at all! $\endgroup$ – zipirovich Aug 8 '17 at 13:44
  • $\begingroup$ sorry guys accidentally accepted the answer, please don't dislike his answer he was guessing $\endgroup$ – Dhruva Aug 8 '17 at 13:45
  • $\begingroup$ @MANMAID Could you please cancel the downvote. I have corrected my proof $\endgroup$ – Raffaele Aug 8 '17 at 14:00
  • $\begingroup$ @zipirovich Could you please cancel the downvote. I have corrected my proof $\endgroup$ – Raffaele Aug 8 '17 at 14:00
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$P(x)=xQ(x)+P(0)$.

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 $P(0)=0\space\space\space\space\blacksquare$

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  • $\begingroup$ I guess no reason to write same answer again and again, but as I started, I posted... $\endgroup$ – MAN-MADE Aug 8 '17 at 14:05

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