Conditional positivity conjecture for $Q_{01}$

Less than 1 year old · traced to 2026

Let $p$ and $q$ be the parameters defining the polynomial $Q_{01}$ . A polynomial has nonnegative coefficients when every coefficient in its expansion is at least zero.

Conditional positivity conjecture. $Q_{01}$ has nonnegative coefficients if and only if $p=2$ and $q\equiv 1 \mod 4$ .

This gives the precise parameter condition for positivity in the $P_{21}$ and $Q_{01}$ case. The supplied text does not state whether the claim has been proved or remains open.

References

Primary source

Evelyn Fiore, George D. Nasr and Cooper Stone, “Dice Relabeling Using Square-Sided Dice”, arXiv:2606.20311 (2026).

generating-functions polynomial-positivity

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written argument claims to prove the conjecture completely, but no independent verification has been found.

The conjecture predicts that nonnegative coefficients occur exactly when $p=2$ and $q\equiv1\pmod4$ . It appears in the June 2026 paper by Evelyn Fiore, George D. Nasr, and Cooper Stone, which presents such statements as directions for future work.

Posted attempt

A reader-written argument claims a complete proof: it handles every odd prime $p$ by exhibiting a negative coefficient and combines this with the asserted $p=2$ classification. The argument has not been independently verified.

Current status (as of August 2026): a complete classification has been claimed in an unverified posted argument, while independent confirmation of the conjecture is absent.

Sources

arxiv.org

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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. \;}