Negative coefficient in an almost cyclotomic polynomial

Question

Let $a,b,c,d$ be four prime numbers. We set the polynomial : $$P(X)=\frac{(1-X^{abc})(1-X^{abd})(1-X^{acd})(1-X^{bcd})(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})(1-X^{ac})(1-X^{ad})(1-X^{bc})(1-X^{bd})(1-X^{cd})}$$ By numerical tests, i see that $P(X)$ always has at least one negative coefficient, how can i prove it?

Answer 1

Suppose that $a<b<c<d$. We show that the coefficient of $X^c$ or of $X^{b+c-1}$ of $P(X)$ is negative. In order to do so, it suffices to work in the power series ring $\mathbb Q[[X]]$ modulo $X^{b+c}$. Note that $ac>b+c$ and so on, hence \begin{equation} P(X)\equiv\frac{(1-X^a)(1-X^b)(1-X^c)(1-X^d)}{(1-X)^2(1-X^{ab})}\pmod{X^{b+c}}. \end{equation} Set \begin{equation} F(X)=\frac{1-X^a}{1-X}\cdot\frac{1-X^b}{1-X}\cdot\frac{1}{1-X^{ab}}=(1+\dots+X^{a-1})(1+\dots+X^{b-1})(1+X^{ab}+\dots), \end{equation} so $P(X)\equiv F(X)(1-X^c-X^d)$ (recall that $c+d>b+c$).

For a power series $G$ let $G[k]$ be the coefficient of $X^k$.

So $P[k]=F[k]-F[k-c]-F[k-d]$.

The coefficients of $F$ lie between $0$ and $a$, and $F[k]=a$ if an only if the remainder of $k\pmod{ab}$ lies between $a-1$ and $b-1$. Furthermore, $F[k]=0$ if the remainder of $k\pmod{ab}$ lies between $a+b-1$ and $ab-1$.

Now assume that $P[b+c-1]\ge0$. From $P[b+c-1]=F[b+c-1]-F[b-1]-F[b+c-1-d]$ and $F[b-1]=a$ we infer that $F[b+c-1]=a$. Thus the remainder of $(b+c-1)\pmod{ab}$ lies between $a-1$ and $b-1$. This implies that the remainder of $c\pmod{ab}$, call it $r$, fulfills $ab+a-b\le r\le ab-1$ or $r=0$. The latter cannot happen, because then $a$ would divide the prime $c$. Thus the former holds. But $a+b-1\le ab+a-b$, hence $F[c]=0$.

But then $P[c]=F[c]-F[0]=0-1=-1$, and we are done.