Louboutin's polynomial identity conjecture for S and R

From papers

For d1d\geq 1, define

Pd(X,Y)=dk,l002k+3ld(1)k1(k+lk)(dk2lk+l)XkYd2k3ldk2l.P_d(X,Y)=d\sum_{\substack{k,l\geq 0\\0\leq 2k+3l\leq d}}(-1)^{k-1}\binom{k+l}{k}\binom{d-k-2l}{k+l}\frac{X^kY^{d-2k-3l}}{d-k-2l}.

Let a,bZa,b\in\mathbb{Z} be nonzero, put c:=a+b0c:=a+b\neq 0, and define

Sa,b(T)=Ta+Tb+Ta+b.S_{a,b}(T)=T^{-a}+T^{-b}+T^{a+b}.

For d{a,b,c}d\in\{a,b,c\},

Pd(Sa,b(T),Sa,b(1/T))=Sa,b(1/Td).P_{|d|}(S_{a,b}(T),S_{a,b}(1/T))=-S_{a,b}(1/T^{|d|}).

If aa is even and bb is odd, also define

Ra,b(T)=Ta(1)a+bTb+Ta+b.R_{a,b}(T)=T^{-a}-(-1)^{a+b}T^{-b}+T^{a+b}.

Then

Pd(Ra,b(T),Ra,b(1/T))={Sa,b(1/Td),d=a,Ra,b(1/Td),d{b,c}.P_{|d|}(-R_{a,b}(T),-R_{a,b}(1/T))=\begin{cases}-S_{a,b}(1/T^{|d|}),&d=a,\R_{a,b}(1/T^{|d|}),&d\in\{b,c\}.\end{cases}

This identity is one of Louboutin's conjectures used in the proof of the paper's main theorem; its resolution status is not specified in the supplied text.

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Sources & referencesView supporting material

Primary source

Jinwoo Choi and Dohyeong Kim, “On a weak form of Ennola's conjecture about certain cubic number fields”, arXiv:2410.21158 (2024).

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