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Suppose that monic polynomial $f(x)\in\Bbb Z[x]$ such that for all $m\in\Bbb Z$, $m>1$, there's no integers $\langle r,r_1,\ldots,r_m\rangle$ such that $f(r)=f(r_1)\cdots f(r_m)$. Is there any interesting property of $f$, or criterion for $f$, or any background?

There's something trivial:

  1. if $\deg f=2$, there's no such polynomial. Let $f(x)=(x+\alpha)(x+\beta)=x^2+bx+c$, we have \begin{align*} f(t)f(t+1) &=(t+\alpha)(t+\beta)(t+1+\alpha)(t+1+\beta)\\ &=(t+\alpha)(t+1+\beta)\cdot (t+1+\alpha)(t+\beta)\\ &=\big(t^2+t+t(\alpha+\beta)+\alpha\beta+\alpha\big)\big(t^2+t+t(\alpha+\beta)+\alpha\beta+\beta\big)\\ &=\big(t^2+(b+1)t+c+\alpha\big)\big(t^2+(b+1)t+c+\beta\big)\\ &=f\big(t^2+(b+1)t+c\big) \end{align*}
  2. $f(x)=(x+1)\cdots(x+l)+2$ satisfies whenever $l\ge4$, for $f(r)\equiv2\pmod4$ whenever $r\in\Bbb Z$, and $f(r_1)\cdots f(r_m)\equiv0\pmod4$.

Could you give me some reference or idea about the criterion of such $f(x)$?

Source Chinese Second Round Olympiad

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    I don't understand "How about the polynomial $f(x)$?". Are you asking whether there exists such an $f$?2012-07-05
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    In your 2nd example, where you write, $f(r)\equiv2\pmod4$ whenever $r$ is in $\bf Z$," that's certainly not true.2012-07-05
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    The problem you have stated is very different from the problem stated at that link.2012-07-05
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    @GerryMyerson edited, $+2$ added.2012-07-05
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    @AlexBecker Not only the existant (when $\deg f\ge4$ is solved) but also the property or criterion.2012-07-05
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    @GerryMyerson You said that my problem is *very different* from the problem in that website, because property (2) stated [there](http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2490476&sid=5b77223a3917658ff65104506205cda1#p2490476) is very different from mine. Am I right? Well, it's because the translate in that website is *WRONG*. I took part in that competition, and we finally found that it's unnecessary that $r_1,\ldots,r_m$ are *distinct*.2012-07-05
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    No, the difference is that you say, no integers such that $f(r)$ equals that product, while at the link it says there exist integers such that $f(r)$ doesn't equal that product.2012-07-05
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    There are lots of examples like #2, e.g., $f(x)=(x+1)\cdots(x+t)+3$ for $t\ge6$, by considerations modulo 9.2012-07-05
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    @GerryMyerson I want to know the background of this problem, and the criteron for such functions.2012-07-06
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    Perhaps you could write to the organizers of the Chinese Olympiad to ask them for the background (I assume you mean, history) of the problem. They might even be able to tell you about the general solution. All I can think of is functions that are always $p^r$ modulo $p^s$ for some prime $p$ and some $r\ge s/2$.2012-07-06
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    @GerryMyerson It's impossible because I cannot know the author of that problem. Exactly, it was not Chinese Mathematics Olympiad. It was only a pre-test for CMO and as a result, I was out.2012-07-06
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    By the way, your equation for degree 2 is wrong.2012-07-06
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    @GerryMyerson Edited. Is it right now? I've checked it with GNU maxima.2012-07-06
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    There's something missing in a step in the middle, but the final result is OK.2012-07-06
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    @GerryMyerson I feel too embarrassed that I was wrong again and again. Now, I've edited with a bug fixed. Any more improvement?2012-07-06

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