0
$\begingroup$

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?

  • 2
    @NoahSnyder - see the meta: http://meta.math.stackexchange.com/questions/6422/a-re-post-and-a-delete-of-the-original-question2019-02-05

1 Answers 1

5

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.

  • 1
    @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