# 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.

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

## 3 Answers

$$\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

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, $$P(-2015)=0$$

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

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 $$b=2015$$.