Conditional positivity conjecture for $Q_{01}$

Proof of the full conditional-positivity conjecture

The source already proves the $p=2$ classification in Theorem 5.8. It remains to prove that $Q_{01}$ has a negative coefficient for every odd prime $p$ and every distinct prime $q$.

Write

$$p=2h+1,\qquad P=p^2,\qquad T(x)=\Phi_q(x)^2,\qquad \Delta(x)=(1-x^p)T(x).$$

The coefficients of $T$ are

$$t_j= \begin{cases} j+1,&0\le j<q,\\ 2q-1-j,&q\le j\le2q-2,\\ 0,&\text{otherwise}. \end{cases}$$

The exact cyclotomic identity

$$\Phi_{p^2q}(x) =\frac{(1-x^{p^2q})(1-x^p)} {(1-x^{p^2})(1-x^{pq})}$$

gives

$$\frac{Q_{01}(x)}x =\frac{\Delta(x)(1-x^{Pq})} {(1-x^P)(1-x^{pq})}.$$

Therefore, whenever $0\le k<pq$,

$$[x^k]\frac{Q_{01}(x)}x =\sum_{j\ge0}\Delta_{k-jP}, \qquad \Delta_i=t_i-t_{i-p}. \tag{1}$$

If $q\le h$, take $k=p$. Since $2q-2<p$,

$$[x^p](Q_{01}/x)=t_p-t_0=-1.$$

If $h<q<P-h$, take $k=q+h<P$. Then

$$[x^{q+h}](Q_{01}/x) =t_{q+h}-t_{q+h-p} =(q-h-1)-(q-h)=-1.$$

This remains valid at $q=h+1$.

It remains to consider $q\ge P-h$. The support of $\Delta$ lies in

$$[0,M],\qquad M=2q+p-2.$$

The complete block

$$M\le k\le M+P-1 \tag{2}$$

lies below $pq$: for $p\ge5$,

$$(p-2)q\ge(p-2)(p^2-p+1) =p^2+p-2+p(p^2-4p+2) \ge p^2+p-2.$$

For $p=3$, the assumption $q\ge8$ forces the distinct prime $q\ge11$, and $M+P-1=2q+9<3q$.

For $r\in\{0,\ldots,P-1\}$, define

$$L_r=\sum_{\substack{0\le i\le M\i\equiv r\pmod P}}\Delta_i.$$

By (1), the coefficients in block (2) are exactly the $P$ numbers $L_r$, each occurring once. Their sum is

$$\sum_{r=0}^{P-1}L_r=\Delta(1)=0.$$

If all these coefficients were nonnegative, every $L_r$ would vanish. Reducing $\Delta(x)$ modulo $x^P-1$ would then imply

$$x^{p^2}-1\mid(1-x^p)\Phi_q(x)^2.$$

Since

$$x^{p^2}-1=(x^p-1)\Phi_{p^2}(x),$$

cancellation would give

$$\Phi_{p^2}(x)\mid\Phi_q(x)^2,$$

contradicting the coprimality of distinct cyclotomic polynomials. Therefore at least one coefficient in (2) is strictly negative.

This covers every odd prime $p$. Combining it with the already proved $p=2$ case of Theorem 5.8 establishes the complete classification

$$\boxed{\; Q_{01}(x)\in\mathbb Z_{\ge0}[x] \quad\Longleftrightarrow\quad p=2\ \text{and}\ q\equiv1\pmod4. \;}$$