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Let $f(x)=a_0+a_1x+ \ldots +a_nx^n$ be a polynomial with integer coefficients, where $a_n>0$ and $n \ge 1$. Prove that $f(x)$ is composite for infinitely many integers $x$.

Proving a polynomial $f(x)$ composite for infinitely many $x$

Can anyone explain why in answer to this question( in the link given above ) such an $m$ always exist such that $f(m)\neq\pm 1$?

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    It's $f(m) \neq \pm 1$. And such an $m$ exists because $\lim\limits_{x\to\infty} \lvert f(x)\rvert = +\infty$ if $\deg f > 0$. – Daniel Fischer Feb 22 '16 at 21:23
  • Fixed that , thanks for mentioning. – User Feb 22 '16 at 21:25
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    @DanielFischer, Is there any other method to show this to a person who don't know about limits ? – User Feb 22 '16 at 21:28
  • Suppose no such $m$ existed, then $f(m) = \pm 1$ for all $m$. If $f(m) = 1$ or $-1$ for all $m$, then it is constant which contradicts the assumption that $f$ isn't constant. If however it achieves both $1$ and $-1$, by the IVT, it must achieve all values in between which contradicts the assumption that $f(m) = \pm 1$ for all $m$. – Cameron Williams Feb 22 '16 at 21:28
  • @CameronWilliams, okie in that case $0$ will be the required integer, as the only integer between -1 and 1 is 0. – User Feb 22 '16 at 21:31
  • @CameronWilliams, but we need $f(m)$ to be an integer – User Feb 22 '16 at 21:33
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    @CameronWilliams $m$ is an integer in that answer, one is only interested in the values at integer arguments. – Daniel Fischer Feb 22 '16 at 21:37
  • @CameronWilliams, yeah that is fine, basically if $f(m)=1 or -1$ for all $m$ then $f$ must also attains $0$ somewhere, and in this case also we can find a prime $p$ such that $p | f(m)$ – User Feb 22 '16 at 21:37
  • @DanielFischer Oh it sure is. I didn't notice the slight difference between the title and question in the OP! – Cameron Williams Feb 22 '16 at 21:38
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    @Member Without limits: If $f(n) \in {1,-1}$ for all $n\in \mathbb{Z}$, then at least one of $1$ and $-1$ is attained infinitely often. But then $f(x) \mp 1$ is a polynomial with infinitely many zeros, but that contradicts $\deg f \geqslant 1$. – Daniel Fischer Feb 22 '16 at 21:39
  • @DanielFischer, thanks it was a nice observation. – User Feb 22 '16 at 21:40
  • @DanielFischer, Can't we also complete the argument given by Cameron Williams. – User Feb 22 '16 at 21:41
  • @Member I think Daniel's approach is a bit better since it doesn't rely on the intermediate value theorem (which is a bit limit-oriented). – Cameron Williams Feb 22 '16 at 21:42
  • @CameronWilliams, Yeah I agree his idea was beautiful but I also want to see can't we do this by your approach. – User Feb 22 '16 at 21:44
  • @Member Actually my suggestion doesn't work as-is since I did not realize that $m$ was taken to be integral because of the mismatch of your title and question initially. Using the IVT forces you to look at all points between two points, but since you are only interested in integer points, I don't think it works out. – Cameron Williams Feb 22 '16 at 21:51

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