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Let $$f(x)=\sum_{i=0}^n a_nx^n$$ be a polynomial with $$a_n \in Z,n>0,a_n\neq0$$ Prove that there exists some natural number $$m>2010$$ such that $$|f(m)|$$ is not a prime number.

I tried to look at f(m) mod m, and i assumed m is relatively prime to a0, so f(m) is reversible mod m. but I didn't know how to continue to prove it is not a prime number.

I would like to get help with that. Thanks


marked as duplicate by Rory Daulton, user147263, Joe Johnson 126, mathlove, Rolf Hoyer Apr 19 '15 at 0:04

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  • $\begingroup$ You might find this useful. Also, I think you meant $a_ix^i$ inside that $\sum$ there. $\endgroup$ – barak manos Apr 18 '15 at 14:52
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    $\begingroup$ In fact, you can probably just choose $m=|2011\cdot\prod\limits_{k=0}^{n}a_k|$. $\endgroup$ – barak manos Apr 18 '15 at 14:54
  • $\begingroup$ @barakmanos Provided that no $a_k$ is zero... But yes, your choice seems simply right $\endgroup$ – mathifold.org Apr 18 '15 at 15:00
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    $\begingroup$ What if $a_0=\pm1$? $\endgroup$ – Empy2 Apr 18 '15 at 15:00
  • $\begingroup$ @Michael: With my suggestion, $m$ will still be larger than $2010$. $\endgroup$ – barak manos Apr 18 '15 at 15:05

Just take $m$ to be some multiple of $a_0$. Then everything will be multiple of $a_0$. Perhaps we have bad luck and $|f(m)|$ is in fact $a_0$... but it cannot happen for all the multiples of $m$ greater than 2010: simply consider that $f(x)=a_0+xg(x)$ with $g$ polynomial; if $x=m$ is $ka_0$, then $f(m)=(kg(ka_0)+1)a_0$, and $kg(ka_0)+1$ won't be $0$ or $-2$ for all the values of $k$ large enough.

  • $\begingroup$ As Michael pointed out, this answer is not valid if $a_0=\pm 1$ :-( $\endgroup$ – mathifold.org Apr 18 '15 at 15:52

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