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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.

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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
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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 $|a_n|\le 2^{n-1}|a_{n-2}|+...$

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