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?
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1$\begingroup$ Related question I asked at mathoverflow.net/questions/214784/… $\endgroup$– Louis DeaettCommented Sep 3, 2015 at 22:17
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2$\begingroup$ Note that $P(x)=\prod \Phi_m(x)$, where the product is over all $1<m<abcd$ dividing $abcd$. $\endgroup$– Louis DeaettCommented Sep 3, 2015 at 22:46
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$\begingroup$ @Louis the references you give seem really interesting... $\endgroup$– Flo140Commented Sep 4, 2015 at 12:42
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$\begingroup$ When we write $P(X)=\prod \phi_{m}(X)$ it's not obvious for me that we can deduce the sign of the coefficient of $P$. For exemple if we consider $$Q(X)=\frac{(1-X^{ab})(1-X^{ac})(1-X^{bc})}{(1-X^{a})(1-X^{b})(1-X^{c})}$$ we have also $Q(X)=\prod \phi_{m}(X)$ but in this case it's clear that $Q$ only has non negative coefficient. $\endgroup$– Flo140Commented Sep 4, 2015 at 13:14
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$\begingroup$ Yes, I certainly agree with you. My observation above is well short of answering the question, sadly. I'd also like to see how this could be resolved. I'd also be interested in knowing how this question came up to begin with, if it's at all easy to explain. $\endgroup$– Louis DeaettCommented Sep 4, 2015 at 13:50
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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.
answered Sep 4, 2015 at 15:32
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1$\begingroup$ Another choice of the exponent which works is the minimal $k\geq c$ such that $(k\mod ab)\geq a+b-1$. $\endgroup$ Commented Sep 5, 2015 at 23:39