Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n\ge 4\in\mathbb N$. Suppose that $a_1,a_2,\cdots,a_{n-1}$ are given integers.

Then, here is my question.

Question : Is the following true for any $(a_1,a_2,\cdots,a_{n-1})$ ?

There exists a composite $a_n$ such that $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n$ cannot be factored using integer coefficients.

Motivation : I've known the following theorem :

Theorem : There exit such prime number $a_n$ for any $(a_1,a_2,\cdots,a_{n-1})$.

Then, I got interested in the case that $a_n$ is not a prime number. The answer must be YES, but I can't prove that. Can anyone help?

In the following, I'm going to write the proof for the above theorem.

Proof for theorem : Let $a_n$ be a prime number. If $f(x)$ can be factored, then we can write $$\begin{align}f(x)&=x^n+a_1x^{n-1}+\cdots+a_n\\ &=(x^m+b_1x^{m-1}+\cdots+b_m)\cdot (x^{n-m}+c_1x^{n-m-1}+\cdots+c_{n-m})\\ &=g(x)\cdot h(x)\ \ \ \ \ (|b_m|\le|c_{n-m}|)\end{align}$$ Since $b_m\cdot c_{n-m}=a_n=\text{a prime number}$, we know that $b_m=\pm 1.$ Letting $\alpha_1,\alpha_2,\cdots,\alpha_m$ be the solutions of $g(x)=0$, since we can write $g(x)=(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_m),$ we know that $(-1)^m\alpha_1\alpha_2\cdots\alpha_m=b_m.$ If $|\alpha_1|\gt 1,|\alpha_2|\gt 1,\cdots,|\alpha_m|\gt 1$, then $|b_m|=|\alpha_1||\alpha_2|\cdots |\alpha_m|\gt 1,$ which is a contradiction. Hence, we know that there exists $\alpha$ such that $f(\alpha)=0, |\alpha|\lt 1$. Hence, we get $$\begin{align}|a_n|&=|{\alpha}^n+a_1{\alpha}^{n-1}+\cdots+a_{n-1}\alpha|\\ & \le |\alpha|^n+|a_1|\cdot |\alpha|^{n-1}+\cdots+|a_{n-1}|\cdot |\alpha|\\ &\le 1+|a_1|+\cdots+|a_{n-1}|\end{align}$$ Hence, we can take a prime number which is larger than $1+|a_1|+\cdots+|a_{n-1}|$ as $a_n$ in order that $f(x)$ cannot be factored using integer coefficients. We now know that the proof is completed.

share|cite|improve this question
I suppose your question really is, “There exists a composite $a_n$ such that ...” – Ewan Delanoy Nov 10 '13 at 15:12
@EwanDelanoy: Thanks. You are right. I edited it. – mathlove Nov 10 '13 at 15:15
up vote 1 down vote accepted

Well, you can easily modify your proof to get a composite $\alpha.$ The only difference you will get is that you need to take $|a_n|$ lager. Namely, take $a_n=2p$ where $p$ is very large prime and run the same proof. Instead of choosing root $|\alpha|<1$ you will be able to choose the root smaller than $2$ and get different bound on $|a_n|\le 2^{n-1}|a_{n-2}|+...$

share|cite|improve this answer
You are right. I haven't noticed this idea. Thanks! – mathlove Nov 11 '13 at 4:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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