Is the following statement true?



> 
> ($\star$) Given integers $n > k > 0$, there exists a monic polynomial of degree $n$ with integer coefficients and constant term $\pm 1$, irreducible over $\mathbb{Z}[x]$, whose roots are all real and simple, with exactly $k$ roots of absolute value $<1$ and $n-k$ roots of absolute value $>1$.
> 
> 
> 


If $k=n-1$ then the answer is yes: in [this old paper by Vijayaraghavan](https://doi.org/10.1017/S0305004100017989) there is an explicit example of such a polynomial (see the first proof of Theorem 1; detailed calculations are omitted). It seems possible to tweak Vijayaraghavan's construction to other values of $k$, but the calculations are somewhat ugly. 


I wonder whether statement ($\star$) has a cleaner proof and/or is already known. 


Remark: If $k=n-1$ then the unique root of absolute value $>1$ is a [Pisot–Vijayaraghavan number](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number). 



Existence of algebraic integers with certain properties
You can even construct such a polynomial for any totally real field $K$ of degree $n$. By Dirichlet's theorem, the unit group $U$ of $K$ maps to


$$U \rightarrow (K^\* \otimes \mathbf{R}) \stackrel{\log| \cdot |}{\longrightarrow}
\mathbf{R}^{n},$$


and the image is a lattice $L$ (of rank $n-1$) in the co-dimension one subspace of elements which sum to zero.
 In particular, such a lattice always contains elements in any orthant corresponding to the first $k$ entries being negative and the next $n-k$ being positive for $k \ne 0,n$. If $\lambda \in L$ is any such element then the corresponding $u \in U$ will have a minimal polynomial of the required form, up to the irreducibility condition.


It remains to show that one can also choose such units which generate the field $K$. But this is easy, because the units which lie in the (finitely many) proper subfields of $K$ give rise to subgroups of the unit group of smaller rank. (This only uses that $K$ is not a totally imaginary CM extension). And certainly $L$ minus the corresponding subgroups will still have many elements in the appropriate orthant. Replacing the orthants by cones one can extend this argument to find units with even more restrictive conditions. 


