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?

  • $\begingroup$ What is the simplest non-constant polynomial? $\endgroup$ – Max Oct 24 '12 at 10:59
  • 1
    $\begingroup$ possible duplicate of Showing $f(x)$ is constant. $\endgroup$ – Belgi Oct 24 '12 at 11:01
  • 2
    $\begingroup$ How does it go if $a_0=\pm 1$? $\endgroup$ – Berci Oct 24 '12 at 11:09
  • 3
    $\begingroup$ Why was the old question deleted? Please undelete it. $\endgroup$ – Noah Snyder Oct 24 '12 at 12:00
  • 2
    $\begingroup$ @NoahSnyder - see the meta: meta.math.stackexchange.com/questions/6422/… $\endgroup$ – 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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.