# Showing $f(x)$ is constant.

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?

• What is the simplest non-constant polynomial? – Max Oct 24 '12 at 10:59
• possible duplicate of Showing $f(x)$ is constant. – Belgi Oct 24 '12 at 11:01
• How does it go if $a_0=\pm 1$? – Berci Oct 24 '12 at 11:09
• Why was the old question deleted? Please undelete it. – Noah Snyder Oct 24 '12 at 12:00
• @NoahSnyder - see the meta: meta.math.stackexchange.com/questions/6422/… – Belgi Oct 24 '12 at 12:27

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.
• 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. – Jack D'Aurizio Oct 25 '12 at 9:29
• But @JackD'Aurizio how does that imply $q\mid f(p+mq)$ for all $m\in\mathbb{N}$? – MickG May 9 '15 at 18:45
• @MickG: $$q\mid mq = (mq+p)-p\mid f(p+mq)-f(p),$$ but since $q\mid f(p)$ we have $q\mid f(p+mq)$. – Jack D'Aurizio May 9 '15 at 20:26