# A coefficient proof of the Bui–Hall sign conjecture

2026-09-13. Candidate full proof; independent review and later-literature comparison pending. This concerns a specific conjecture about the leading constants in mixed fourth moments of Hardy's Z-function. It proves neither RH nor the cycle19 defect estimate G. No originality or formal-verification claim is made here.

## Exact external statement

Let alpha=(k,l,m,n) be any four nonnegative integers, S=k+l+m+n, and r the number of odd entries. Bui–Hall, [On the derivatives of Hardy's function Z(t), BLMS 55 (2023), 2304–2323](https://londmathsoc.onlinelibrary.wiley.com/doi/full/10.1112/blms.12859), Theorem 3, gives

\[
\operatorname{HARDY}(k,l,m,n)=3(-1)^{m+n}i^S I_\alpha,
\tag{1}
\]

where

\[
I_\alpha=\int_{[0,1]^4}(u_1-u_2)^2 A^k B^l C^m D^n\,d\mathbf u,
\]

\[
A=\tfrac12-u_1+(u_1-u_2)u_3,\quad
B=\tfrac12-u_2-(u_1-u_2)u_3,
\]
\[
C=\tfrac12-u_1+(u_1-u_2)u_4,\quad
D=\tfrac12-u_2-(u_1-u_2)u_4.
\]

The same formula and Conjecture 1 were directly inspected in [arXiv:2304.05178v1](https://arxiv.org/html/2304.05178v1). The source defines HARDY as pi squared times the coefficient of T(log T)^(S+4) in the mixed fourth moment. We use its moment theorem as an external published input, without re-proving it.

**Theorem.** For every alpha of even total degree,

\[
(-1)^{\lfloor k/2\rfloor+\lfloor l/2\rfloor+\lfloor m/2\rfloor+\lfloor n/2\rfloor}
\operatorname{HARDY}(k,l,m,n)>0.
\tag{2}
\]

Thus HARDY is nonzero, and is negative exactly when |i^k+i^l+i^m+i^n|=2, as in Conjecture 1. Odd total degree gives zero, consistently with the source's Corollary 1.

## 1. Exact reduction to three independent Laplace variables

Put x=1/2-u1 and y=1/2-u2. For any a,b define the oriented integral

\[
J_{a,b}(x,y)=\int_x^y t^a(x+y-t)^b\,dt.
\]

The changes in u3 and u4 each divide by y-x, and cancel the original (y-x)^2. The diagonal x=y has measure zero. Hence

\[
I_\alpha=\int_{[-1/2,1/2]^2}J_{k,l}(x,y)J_{m,n}(x,y)\,dx\,dy.
\tag{3}
\]

The product is unchanged by exchanging x,y. Restrict to x<=y and multiply by two. Write s=x+y, d=y-x. The Jacobian is 1/2 and the range is |s|<=1, 0<=d<=1-|s|. In the two inner integrals put t=(s+v)/2 and t'=(s+w)/2, so their combined Jacobian is 1/4 and |v|,|w|<=d. Integrating d first gives

\[
I_\alpha=\frac14\int_{|s|+\max(|v|,|w|)\le1}
(1-|s|-\max(|v|,|w|))
\left(\frac{s+v}{2}\right)^k\left(\frac{s-v}{2}\right)^l
\left(\frac{s+w}{2}\right)^m\left(\frac{s-w}{2}\right)^n,ds\,dv\,dw.
\tag{4}
\]

Set u=s, z=(v-w)/2, h=(v+w)/2. Then dv dw=2 dz dh and max(|v|,|w|)=|z|+|h|. Define the four unscaled forms

\[
L_1=u+z+h,\quad L_2=u-z-h,\quad
L_3=-u+z-h,\quad L_4=-u-z+h,
\]

and P_alpha=L1^k L2^l L3^m L4^n. With rho=|u|+|z|+|h|, (4) becomes

\[
I_\alpha=(-1)^{m+n}2^{-S-1}
\int_{\rho\le1}(1-\rho)P_\alpha(u,z,h)\,du\,dz\,dh.
\tag{5}
\]

For any homogeneous polynomial P of degree S in three variables, integration separately on the eight orthants and then in the radial variable rho gives

\[
\int_{\rho\le1}(1-\rho)P
=\frac1{(S+4)!}\int_{\mathbb R^3}e^{-\rho}P.
\tag{6}
\]

Indeed the radial factors are respectively 1/((S+3)(S+4)) and (S+2)!; signed angular integrals cause no difficulty. All integrals are absolutely convergent. Let U,Z,H be independent real variables of density e^(-|t|)/2, and let X_i=L_i(U,Z,H). Equations (1),(5),(6) yield, for even S,

\[
\boxed{\operatorname{HARDY}(\alpha)
=\frac{12(-1)^{S/2}}{2^S(S+4)!}
\mathbb E[X_1^kX_2^lX_3^mX_4^n].}
\tag{7}
\]

The constant gives HARDY(0,0,0,0)=1/2. For odd S the expectation vanishes by simultaneous reflection.

## 2. Parity transform of the generating function

All power series below may be read formally over the rationals, or analytically near zero. Write

\[
p_0=t_1+t_2+t_3+t_4,\quad p_1=t_1+t_2-t_3-t_4,
\]
\[
p_2=t_1-t_2+t_3-t_4,\quad p_3=t_1-t_2-t_3+t_4,
\quad A_j=1-p_j^2.
\]

The Laplace moment generating function is

\[
M(\mathbf t)=\mathbb E e^{\sum t_iX_i}=\frac1{A_1A_2A_3}.
\tag{8}
\]

For a monomial of exponent alpha, multiply its coefficient in M by i^r, where r is the number of odd entries of alpha. Since odd total degrees in M vanish, the transformed series G is real. Averaging over coordinate sign changes gives

\[
G=\tfrac12\{M(t_1,t_2,-t_3,-t_4)+M(t_1,-t_2,t_3,-t_4)
+M(t_1,-t_2,-t_3,t_4)-M(\mathbf t)\}.
\tag{9}
\]

For example this follows from 2^(-4) sum_epsilon product_j(1+i epsilon_j) M(epsilon_1 t1,...,epsilon_4 t4); terms with an odd number of negative signs have purely imaginary weights and cancel in antipodal pairs. Multiplying out the four denominators in (9) gives

\[
\boxed{G=\frac{1-\sum_jt_j^2+2\sum_{i<j}t_it_j}
{\prod_{j=0}^3(1-p_j^2)}=(1-Q)R+2\sum_{i<j}t_it_jR,}
\tag{10}
\]

where Q=sum t_j^2 and R=1/product A_j. This uses sum_(j=0)^3 p_j^2=4Q.

## 3. Strict positivity of every even-degree coefficient

Let E be the eight sign vectors epsilon in {+1,-1}^4 with product epsilon_j=1. Factoring each 1-p_j^2 shows

\[
\log R=\sum_{d\ge1}\frac1d\sum_{\epsilon\in E}(\epsilon\cdot\mathbf t)^d.
\tag{11}
\]

The character sum sum_E epsilon^alpha is eight if the four entries of alpha are all even or all odd, and zero otherwise. (These are exactly the characters trivial on E.) Thus log R has nonnegative coefficients and zero constant term. For |alpha|=2N>0,

\[
[\mathbf t^\alpha]\log R=
\begin{cases}\dfrac4N\binom{2N}{\alpha_1,\alpha_2,\alpha_3,\alpha_4},&\text{all entries have the same parity},\\
0,&\text{otherwise}.
\end{cases}
\tag{12}
\]

Set F=log R+log(1-Q). The second summand subtracts a coefficient only at all-even alpha=2a with |a|=N, namely (1/N) binomial(N;a). Consequently

\[
[\mathbf t^{2a}]F=\frac1N\left(4\binom{2N}{2a}-\binom N a\right)>0.
\tag{13}
\]

The inequality follows from binomial(2N;2a)>=binomial(N;a): map each word with a_j copies of letter j to the word with every letter duplicated. At every all-odd exponent vector the coefficient of F is the strictly positive coefficient in (12). All other coefficients of F are zero. In particular F is coefficientwise nonnegative, with zero constant term. Therefore

\[
(1-Q)R=\exp F
\tag{14}
\]

is coefficientwise nonnegative, and has strictly positive coefficients at every all-even nonzero or all-odd exponent vector; its constant is one. Similarly R=exp(log R) is coefficientwise nonnegative and positive at every all-even vector, including zero.

Every even-total alpha has r=0,2,or4. For r=0 or4, its coefficient in (14) is strictly positive. For r=2, choose the two odd positions i,j; then alpha-e_i-e_j is nonnegative and all-even, and the term 2t_i t_j R in (10) supplies a strictly positive coefficient. All remaining contributions in (10) are nonnegative. Hence

\[
[\mathbf t^\alpha]G>0\qquad (|\alpha|\text{ even}).
\tag{15}
\]

By the definition of G and the exponential coefficient convention, (15) says (-1)^(r/2) E[X^alpha]>0. Equation (7) proves (2), since sum floor(alpha_j/2)=(S-r)/2 and (S+r)/2 has the same parity.

Finally the equivalence to the source's absolute-value condition is elementary. If r=0 or4, four signs on one coordinate axis sum to a vector of length two exactly when an odd number of their signs are negative. If r=2, the two real and two imaginary signs give length two exactly when one pair has opposite signs and the other equal signs. In both cases this is exactly (-1)^(sum floor(alpha_j/2))=-1. This covers zero indices and all mod-four patterns.

## Trust and remaining decisions

The argument is an elementary integral and formal-series proof for every quadruple, not a finite verification. Its application to zeta uses the external Bui–Hall Theorem 3. An independent source/parity/geometry review and an exact arithmetic cross-check are required before repository acceptance. A later-literature audit is required before an originality claim. No RH, new zero-free region, simple-zero proportion, G, or Lean result follows.