9
$\begingroup$

Let $f(x)=a_0+a_1x+ \ldots +a_nx^n$ be a polynomial with integer coefficients, where $a_n>0$ and $n \ge 1$. Prove that $f(x)$ is composite for infinitely many integers $x$.

I can easily show that there are infinitely many composite numbers of the form $a_0+a_1x+ \ldots +a_nx^n$ if $a_0 \ge 2$, we just note that $f(x)$ is composite for every $x$ being a multiple of $a_0$. But I can't find a way to prove this in the case $a_0=1$.

  • 2
    The question reminded me of [$n^2-n+41$](http://math.stackexchange.com/questions/53686/to-what-extent-can-values-of-n-such-that-n2-n41-is-composite-be-predicted).2011-11-27

2 Answers 2

8

Choose $m$ such that $f(m)\ne\pm1$, then choose any prime $p$ dividing $f(m)$, and think about $f(m+pk)$ for $k=1,2,\dots$.

  • 0
    If anyone is wondering why there exists $m$ with $f(m)\neq \pm 1$, this is the subject of [this question](https://math.stackexchange.com/questions/1667705/how-to-show-that-a-polynomial-fx-is-composite-for-infinitely-many-integers).2018-10-10
-3

Let $x_k=(\text{lcm}(a_0,a_1,\dots,a_n))^k$ for some $k$. Then all $x_k$ with $k\in\mathbb N$ are values which produce composite numbers.

  • 0
    What if $\,f(x)=x^3+x^2+x+1\,$.2018-07-27