TITLE: Can a product of conjugates be a Pisot number again?
QUESTION [7 upvotes]: Let $p(X) \in \mathbb{Z}[X]$ be an irreducible polynomial, and let $\alpha_1 \dots, \alpha_n$ be its roots in $\mathbb{C}$. Suppose that $\alpha_1$ is a Pisot number, that is, $\alpha_1 \in \mathbb{R}, \ |\alpha_1| > 1$, and for any $1 < i \leq n$ we have $|\alpha_i| < 1$.

Suppose that $\alpha_1 < 2$. Is it possible that for some $1 <  k \leq n$ the number $\alpha_1 \cdots \alpha_k$ is Pisot?

Note that the assumption that $\alpha_1 < 2$ implies that $\alpha_1 \in \mathcal{O}_{\mathbb{Q}(\alpha_1)}^*$.

REPLY [3 votes]: Still, it seems to me that the difference in definitions is not that negligible. But if you are OK with negative numbers --- here is an example.
Set $g(x)=x^2-x$ and $f(x)=g(x)^2-g(x)-1=x^4-2x^3+x-1$. Its roots are the roots of
$$
  g(x)=\frac{1\pm \sqrt5}2.
$$
When the '$+$' sign is chosen, we get two real numbers $\alpha_1\in(1,2)$ and $\alpha_2\in(-1,0)$ since $g(-1)=g(2)>\phi$ but $g(0)=g(1)<\phi$. For the '$-$' sign we have two complex roots whose (equal) absolute values are less than 1 since their product's one is such. Thus $\alpha_1$ is a required example, because $\alpha_1\alpha_2=-\phi$. 
Remark. I cannot transfom this example into one where the product is a positive Pisot number, and it looks like much more difficult to make, at least with the same method.