Let B_n be the book graph with parameter n, and let D(B_n,x) denote its domination polynomial. The numerical data suggest the following parity-dependent statement. Real-root conjecture for book graphs. If n is even, then D(B_n,x) has exactly four real roots counting multiplicity: 0 with multiplicity 2, one simple root in (-∞,-2), and one simple root in (-tfrac12,0). If n is odd, then D(B_n,x) has exactly four real roots counting multiplicity: 0 with multiplicity 2 and two simple roots in (-tfrac12,0). The preceding proposition establishes the existence of the two indicated roots in the even case, but the precise complete categorization of real roots remains unresolved.

Parity-dependent real-root conjecture for domination polynomials of book graphs

Less than 1 year old · traced to 2026

Let $B_n$ be the book graph with parameter $n$ , and let $D(B_n,x)$ denote its domination polynomial. The numerical data suggest the following parity-dependent statement. Real-root conjecture for book graphs. If $n$ is even, then $D(B_n,x)$ has exactly four real roots counting multiplicity: $0$ with multiplicity $2$ , one simple root in $(-\infty,-2)$ , and one simple root in $(-\tfrac12,0)$ . If $n$ is odd, then $D(B_n,x)$ has exactly four real roots counting multiplicity: $0$ with multiplicity $2$ and two simple roots in $(-\tfrac12,0)$ . The preceding proposition establishes the existence of the two indicated roots in the even case, but the precise complete categorization of real roots remains unresolved.

References

Primary source

Bilal Ahmad Rather, “On roots of domination polynomials for friendship and book graphs”, arXiv:2604.08998 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The conjecture is supported by a 2026 preprint, but an unverified calculation claims it fails at the smallest odd case, so the general question is unsettled.

Rather’s April 2026 preprint records the parity-dependent root pattern as Conjecture 5.4, based on numerical data, and explicitly leaves the complete classification unresolved. Its formula is $D(B_n,x)=(2x+1)(x^2+2x)^n+x^2(x+1)^{2n}-2x^n$ .

Known results

For even $n\ge2$ , Rather (2026) proves a real root in $(-\infty,-2)$ and one in $(-\tfrac12,0)$ .
Rather (2026) determines the limiting set of domination roots for the book graphs, but this does not classify finite- $n$ real roots.
Numerical data for the tested parameters match the conjectured parity pattern.

Posted attempt

An unverified calculation claims a counterexample at $n=1$ : using the stated formula, $D(B_1,x)=x^2(x^2+4x+6)$ , whose quadratic factor has discriminant $-8$ and hence no real nonzero roots. This would refute the odd case as stated, but it does not address odd $n\ge3$ and has not been independently verified.

Current status (as of August 2026): The even-case existence results and numerical evidence are established, while the full classification remains open; the $n=1$ counterexample claim is unverified.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture is false for the admissible odd parameter $n=1$ .

The source defines the book graph $B_n$ for every $n\ge1$ , and Theorem 2.3 gives

D(B_n,x) =(x^2+2x)^n(2x+1) +x^2(1+x)^{2n} -2x^n.

For $n=1$ , one has $B_1=C_4$ , and the displayed formula becomes

\begin{aligned} D(B_1,x) &=(x^2+2x)(2x+1)+x^2(1+x)^2-2x\\ &=x^2(x^2+4x+6). \end{aligned}

The quadratic factor has discriminant

\Delta=4^2-4\cdot6=-8<0.

Therefore the only real root is $0$ , with multiplicity two. In particular, there are NO real roots in $(-1/2,0)$ , whereas the asserted odd- $n$ case requires two simple real roots in that interval.

Hence Conjecture 5.4(ii), as stated for odd $n$ , is false. Restricting it to odd $n\ge3$ would produce a different corrected conjecture, which is not resolved by this counterexample.