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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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  • $\begingroup$ What is the simplest non-constant polynomial? $\endgroup$ – Max Oct 24 '12 at 10:59
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    $\begingroup$ possible duplicate of Showing $f(x)$ is constant. $\endgroup$ – Belgi Oct 24 '12 at 11:01
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    $\begingroup$ How does it go if $a_0=\pm 1$? $\endgroup$ – Berci Oct 24 '12 at 11:09
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    $\begingroup$ Why was the old question deleted? Please undelete it. $\endgroup$ – Noah Snyder Oct 24 '12 at 12:00
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    $\begingroup$ @NoahSnyder - see the meta: meta.math.stackexchange.com/questions/6422/… $\endgroup$ – Belgi Oct 24 '12 at 12:27
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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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  • $\begingroup$ why step2 holds? $\endgroup$ – Leitingok Oct 25 '12 at 2:06
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Oct 25 '12 at 9:29
  • $\begingroup$ But @JackD'Aurizio how does that imply $q\mid f(p+mq)$ for all $m\in\mathbb{N}$? $\endgroup$ – MickG May 9 '15 at 18:45
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    $\begingroup$ @MickG: $$q\mid mq = (mq+p)-p\mid f(p+mq)-f(p),$$ but since $q\mid f(p)$ we have $q\mid f(p+mq)$. $\endgroup$ – Jack D'Aurizio May 9 '15 at 20:26

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