TITLE: Are there irreducible polynomials with all zeros on two concentric circles?
QUESTION [27 upvotes]: This is somewhat similar to this recent question, but extending in a different direction.
Let $f(x)$ be an irreducible polynomial of degree $n$ with integer coefficients. Call such $f$ a bicycle polynomial if all its roots are located on two circles around $O$, i.e. all roots have one of two moduli. (Of course we'll exclude polynomials of cyclotomic type like $\Phi_n(mx)$ for $m\in\mathbb Z$, which have all their roots fit on one circle.)
For $n=2k$ or $n=3k$, examples of bicycle polynomials are $f(x)=g(x^k-1)$, where $g$ is irreducible of degree $2$ or $3$. Taking here a $g$ of degree $4$ with two pairs of complex roots yields still other bicycle polynomials for $n=4k$.   
Now: except replacing the "$-1$" by any other nonzero integer, those types of constructions already seem to be about all of it... 


Do bicycle polynomials of degree $n$ exist if $n$ has only prime factors $>3$?


Assuming the existence of such a polynomial, it appears (?) to boil down to the existence of a degree $m$ polynomial ($m>3$ odd) with $m-1$ roots on one circle. This circle must presumably have an irrational radius because of what is known about Salem polynomials, but I'm stuck here.  
Also related: How to best distribute points on two concentric circles?
Another question:  


Does anything change if we allow complex integers as coefficients?

REPLY [10 votes]: I couldn't resist to include a picture showing how asymmetric the distribution of roots can be for a bicycle polynomial. Here are the roots of $$z^{16}+2 z^{15}+z^{14}-2 z^{13}-4 z^{12}-2 z^{11}+3 z^{10}\\ +5 z^9+3 z^8+z^7-z^5-z^4-z^3+z^2+z+1,$$ which is an example of an extended Salem polynomial given by Dubickas and Smyth in their paper.