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?
Best Answer
Dubickas and Smyth (On the Remak Height, the Mahler Measure and Conjugate Sets of Algebraic Numbers Lying on Two Circles, 2001) discuss what they call extended Salem numbers.
Moreover, they present results in the direction about which you ask, i.e., conditions under which certain conjugates all fall on two (not one) circles, thereby forcing their associated minimal polynomial to have its degree divisible by either 3 or 2 in parts (a) and (b), respectively.
In the link above, you can find a discussion of extended Salem Numbers and more.