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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 $f$? –  Alex Becker Jul 5 '12 at 3:30
    
In your 2nd example, where you write, $f(r)\equiv2\pmod4$ whenever $r$ is in $\bf Z$," that's certainly not true. –  Gerry Myerson Jul 5 '12 at 4:02
    
The problem you have stated is very different from the problem stated at that link. –  Gerry Myerson Jul 5 '12 at 4:05
    
@GerryMyerson edited, $+2$ added. –  Frank Science Jul 5 '12 at 10:40
    
@AlexBecker Not only the existant (when $\deg f\ge4$ is solved) but also the property or criterion. –  Frank Science Jul 5 '12 at 10:42

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