If $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0$, must we have always $-\frac{a_0}{a_n} \in \mathbb{Z}$?

Question

Let consider the polynomial with integer coefficients: $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ If $f(x)=0$ and $x \in \mathbb{Z}$ with $a_n\neq 0$

If all the roots are integers, must we always have $-\frac{a_0}{a_n} \in \mathbb{Z}$?

Answer 1

Yes. If a polynomial with all integer coefficients has only integer roots then it must be the case that it factors into: $p(x) = c(x-k_1)(x-k_2)(x-k_3)\dots(x-k_n)$ where $c$ and each $k$ are integers. So $a_n= c$ and $a_0 = \pm ck_1k_2k_3\dots k_n$. So $a_n$ divides $a_0$.

Answer 2

Here there is an answer to an older version of the question where we didn't have the assumption that all roots are integer (thanks to berto for the suggestion!)

In other words, are you asking "if $f(x)=a_nx^n+\ldots+a_0 \in \mathbf{Z}[x]$ has a integer root $r$ then $a_n$ divides $a_0$?"

The assumption is equivalent to $0=a_nr^n+\ldots+a_1r+a_0$. There is no such implication.

Counterexample: $f(x)=2x^2-3x+1$ and $f(1)=0$.

Answer 3

Yes. the ratio $-\frac{a_0}{a_n}$ must always be an integer otherwise at least one of the roots is an indivisible fraction. let suppose that all the roots of the polynomial are $x_i \in \mathbb{Z}$ with $1 \leq x \leq n$ Therefore, $$(x-x_1)(x-x_2)\cdots(x-x_n)=0$$ After expansion, we have: $$x_1+x_2+\cdots+x_n=-a_{n-1}$$

$$x_1x_2\cdots x_n=\pm a_0/a_n$$ depending on the parity of $n$ If since $x_i$ is an integer for all $i$ then the product is also an integer.

However, the inverse is not always true. For instance, if that ratio is an integer, it can mean the roots are the inverse of one another.