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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.

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    $\begingroup$ It's a general fact that for integers $a,b$ we have $a-b\mid P(a)-P(b)$. $\endgroup$ – Wojowu Jun 1 '16 at 10:57
  • $\begingroup$ thank you very much..@Wojowu $\endgroup$ – Utkarsh Jun 1 '16 at 11:04
  • $\begingroup$ Note $ \ $ All answers were merged from a different question, so they may not be completely in sync with this question. $\endgroup$ – Bill Dubuque Oct 11 '18 at 16:53
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$\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

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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, $P(-2015)=0$

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  • $\begingroup$ How can you tell that $P(n) divides $P(P(n)-2015)-P(-2015)$... Please elaborate $\endgroup$ – Mayank Mishra Oct 6 '18 at 20:00
  • $\begingroup$ @MayankMishra Suppose if you consider two numbers $x,y$ then $x-y|P(x)-P(y)$ $\endgroup$ – Key Flex Oct 6 '18 at 20:02
  • $\begingroup$ Where can I get this fact explained with proof... Can u provide with a reference or link $\endgroup$ – Mayank Mishra Oct 6 '18 at 20:05
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    $\begingroup$ @MayankMishra It's the Factor Theorem. I added a simpler form in my answer, i.e. subtracting the constant term from a polynomial leaves a polynomial divisible by $x$ ; the constant term is obtained by evaluating the polynomial at $x=0,\,$ e.g. $p(x+a)$ has constant term $p(a)$ $\endgroup$ – Bill Dubuque Oct 6 '18 at 20:21
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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 $b=2015$.

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