# Are there irreducible polynomials with all zeros on two concentric circles?

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.

Another question:

• Does anything change if we allow complex integers as coefficients?
• There are many more examples than the cyclotomic ones that have all roots lying on one circle. In particular, for $\alpha \neq 0$ any element in the union $\mathbb{Q}^{\mathrm{ab}}$ of all cyclotomic fields, the minimum polynomial of $\xi := \bar{\alpha} / \alpha$ will have all its roots on the unit circle; and in general, such a $\xi$ is not a root of unity. Of course, such polynomials have no real roots and hence they all have even degree. But they give other constructions of bicycle polynomials: if $\beta$ is a root of a bicycle polynomial, then so is $\xi \beta$, for any $\xi$ as above. – Vesselin Dimitrov Oct 5 '15 at 3:52
• A wide class of polynomials with all roots on the unit circle is given by the Lee-Yang Circle Theorem. See arxiv.org/pdf/1201.3169v1.pdf. Incidentally, the theorem is named after the two physicists who won a Nobel Prize for the overthrow of parity. – Richard Stanley Oct 5 '15 at 23:32

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.