On Newman polynomials which divide no Littlewood polynomial

Recall that a polynomial $P(x) \in \mathbb{Z}[x]$ with coefficients $0, 1$ and constant term $1$ is called a Newman polynomial, whereas a polynomial with coefficients $-1, 1$ is called a Littlewood polynomial. Is there an algebraic number $\alpha$ which is a root of some Newman polynomial but is not a root of any Littlewood polynomial? In other words (but not equivalently), is there a Newman polynomial which divides no Littlewood polynomial? In this paper, for each Newman polynomial $P$ of degree at most $8,$ we find a Littlewood polynomial divisible by $P$. Moreover, it is shown that every trinomial $1+ux^a+vx^b,$ where $a<b$ are positive integers and $u, v \in \{-1,1\},$ so, in particular, every Newman trinomial $1+x^a+x^b,$ divides some Littlewood polynomial. Nevertheless, we prove that there exist Newman polynomials which divide no Littlewood polynomial, e.g., $x^9+x^6+x^2+x+1.$ This example settles the problem 006:07 posed by the first named author at the 2006 West Coast Number Theory conference. It also shows that the sets of roots of Newman polynomials $V_{\mathcal{N}}$, Littlewood polynomials $V_{\mathcal{L}}$ and $\{-1,0,1\}$ polynomials $V$ are distinct in the sense that between them there are only trivial relations $V_{\mathcal{N}}\subset V$ and $V_{\mathcal{L}}\subset V.$ Moreover, $V \ne V_{\mathcal{L}} \cup V_{\mathcal{N}}.$ The proofs of several main results (after some preparation) are computational.