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Let $f(x)=a_nx^n+...a_1x+a_0$ is an integer polynomial with $a_n>0,n\not=1$. $f(p)$ is prime for every $p$, where $p$ is prime.

How to show $f(x)$ is constant, or not?

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    What is the simplest non-constant polynomial?2012-10-24
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    possible duplicate of [Showing $f(x)$ is constant.](http://math.stackexchange.com/questions/220018/showing-fx-is-constant)2012-10-24
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    @Max: unfortunately $n\ne 1$ is assumed.2012-10-24
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    @Belgi,that's my wrong description.2012-10-24
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    @Leitingok then edit the original question, don't post a new one woth a revised version2012-10-24
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    How does it go if $a_0=\pm 1$?2012-10-24
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    if $a_0$ is not prime then so it $f(0)$...2012-10-24
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    @Berci then $a_0 = p_1k$ where $p_1$ prime,$f(p_1)$ is not prime2012-10-24
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    But, this argument is not that clear.. what about $f(x)=x^2-2$? Then $2|f(2)$ but $f(2)$ is still prime.2012-10-24
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    mhhhhhh, you are right....2012-10-24
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    Why was the old question deleted? Please undelete it.2012-10-24
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    @NoahSnyder - see the meta: http://meta.math.stackexchange.com/questions/6422/a-re-post-and-a-delete-of-the-original-question2012-10-24

1 Answers 1

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Step1. There is at least one prime $p$ for which $f(p)=q\neq p$. Otherwise, the polynomial $f(x)-x$ would have an infinite number of zeros, that implies $f(x)=x$, but the degree of $f$ is different from one.

Step2. In $p(x)$ is a polynomial with integer coefficients and $a,b$ are two different integers, $(a-b)|(p(a)-p(b))$. This implies that $q$ divides $f(p+mq)$ for every natural number $m$.

Step3. By Dirichlet Theorem, there are an infinite number of positive integers $m$ for which $p+mq$ is a prime. Let $M$ be the set of such integers. By the previous step we have: $$\forall m\in M,\quad q\; |\; f(p+mq), $$ but the RHS is a prime, so: $$\forall m\in M,\quad f(p+mq) = q. $$

Step4. By the previous step, we have that $f(x)-q$ has an infinite number of integer roots, so $f(x)$ is constant.

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    why step2 holds?2012-10-25
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    Because, if $a$ and $b$ are different integers, $(a-b)|(p(a)-p(b))$ holds for every $p(x)$ in the form $p(x)=x^k$, so it holds for every polynomial with integer coefficients.2012-10-25
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    But @JackD'Aurizio how does that imply $q\mid f(p+mq)$ for all $m\in\mathbb{N}$?2015-05-09
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    @MickG: $$q\mid mq = (mq+p)-p\mid f(p+mq)-f(p),$$ but since $q\mid f(p)$ we have $q\mid f(p+mq)$.2015-05-09