$f(X) = \sum_{i=0}^{n}a_i X^i$ is a polynomial with $a_i \in\mathbb Z $, $n>0$ and $a_n \neq 0$. Prove that there exists some natural number $m>2016$ such that $|f(m)|$ is not a prime number.

I need hints how to approach this, since no idea has crossed my mind. thank you

  • $\begingroup$ What do you think about $m = |a_{0}| \cdot 2016$? $\endgroup$ – openspace Mar 23 '16 at 10:12
  • $\begingroup$ @openspace what if $a_0 = 1$? $\endgroup$ – crbah Mar 23 '16 at 10:14
  • $\begingroup$ @openspace might not work if $a_0=0$ or 1. $\endgroup$ – BigbearZzz Mar 23 '16 at 10:14
  • $\begingroup$ @BigbearZzz if $a_0=0$, $f(m)$ is divisible by $m$. $\endgroup$ – Darío G Mar 23 '16 at 10:16
  • $\begingroup$ @Wore doesn't change that fact that m = 0*2016 < 2016. $\endgroup$ – BigbearZzz Mar 23 '16 at 10:17

Suppose $|f(2017)|$ is not a prime. Then you are done, pick $m=2017$.

Otherwise, $|f(2017)|=p$ is a prime. Now, all numbers of the form $2017+kp$, with $k \ge 0$, satisfy $$f(2017 + kp) \equiv 0 \pmod{p}$$ and only finitely many of them satisfy $f(2017 + kp) \in \{ 0,p,-p\}$. So, there exist some $k \ge 0$ such that $f(2017+kp)$ is a (proper) multiple of $p$. For such number $k$, $|f(2017+kp)|$ is composite.

  • $\begingroup$ how do you explain that only finitely many satisfy what you wrote? $\endgroup$ – CnR Mar 23 '16 at 10:56
  • $\begingroup$ Since $f$ is a polynomial, the equation $f(x)=p$ has ony finitely many solutions (it could be that there are no solution at all, but, you know, $0$ is a finite number). The same for $f(x)=0$ and $f(x)=-p$. $\endgroup$ – Crostul Mar 23 '16 at 11:13

There is a hint under "Prime formulas and polynomial functions":


  • $\begingroup$ How does this help? $\endgroup$ – Crostul Mar 23 '16 at 10:33
  • $\begingroup$ OP asked for hint. It's good enough to deduce some variation of your answer. $\endgroup$ – zaarcis Mar 25 '16 at 11:52

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