A primitive-ratio cut and a $7/12$ bound for the $abc$ exceptional set

Abstract.

We prove $N_{1}(X)\ll_{\varepsilon}X^{7/12+\varepsilon}$, where $N_{\lambda}(X)$ counts the $abc$ triples of exponent $\lambda$ in $[1,X]^{3}$. The previous bound proved in full was $X^{13/20+\varepsilon}$, and Bernert (arXiv:2506.13364, Section 4) sketched $X^{8/13+\varepsilon}$. We keep the anatomic reduction of Bernert, Browning, Lichtman and Teräväinen. It gives $22$ inequalities for the exponents of a box. We add two exact rows per side for the root classes of index at least seven, and a cut. The cut fixes all but two coprime products on two sides. The remaining pairs are primitive points of a lattice. Its determinant is a product of roots on the third side. This gives $1\,560\,195$ further inequalities. An exact Farkas certificate over $8192$ cells and $10\,820$ leaves gives exponent $7/12$ for the box count. On each leaf the multipliers sum to at most $18/7$. This pays for the losses at $\lambda=1$.

1. Introduction

For this paper $\mathbb{N}=\{1,2,\ldots\}$. For $n\neq 0$ put $\operatorname{rad}(n)=\prod_{p\mid n}p$, and write $\omega(n)=\#\{p:p\mid n\}$, with $\omega(1)=0$. Following [2], call $(a,b,c)\in\mathbb{N}^{3}$ with $\gcd(a,b,c)=1$ an $abc$ triple of exponent $\lambda$ if

(1.1) $$a+b=c,\qquad\operatorname{rad}(abc)<c^{\lambda},$$

Let $N_{\lambda}(X)$ count these triples in $[1,X]^{3}$. The $abc$ conjecture of Masser and Oesterlé says that $N_{\lambda}(X)=O_{\lambda}(1)$ for $\lambda<1$, so $\lambda=1$ is the endpoint of interest.

The bound $N_{\lambda}(X)\ll_{\varepsilon}X^{2\lambda/3+\varepsilon}$ is classical [2, Proposition 1.1]; it comes from de Bruijn’s count [4] of integers with restricted radical. For $\lambda\geq 2$, Kane’s result [6] gives $N_{\lambda}(X)\ll_{\varepsilon}X^{\lambda-1+\varepsilon}$ [2, Section 1]. On $2\leq\lambda\leq 3$ it matches $N_{\lambda}(X)\gg X^{\lambda-1-\varepsilon}$. [2, Section 1] counts it in Kane’s result and states it for every $\lambda>1$. It is originally due to unpublished work of Granville; the question is Mazur’s. It can hold only for $\lambda\leq 3$, since $N_{\lambda}(X)\leq\#\{(a,b,c)\in[1,X]^{3}:a+b=c\}=O(X^{2})$. Browning, Lichtman and Teräväinen [3, Theorem 1.2] gave the first power saving below $2\lambda/3$, for $0.99<\lambda<1.001$. Bernert [1, Theorem 1] proved

(1.2) $$N_{\lambda}(X)\ll_{\varepsilon}X^{(23\lambda+3)/40+\varepsilon}\qquad(0<\lambda<2),$$

so $N_{1}(X)\ll_{\varepsilon}X^{13/20+\varepsilon}$. In the merged paper, [2, Theorem 1.2] states (1.2) for $0<\lambda\leq 2$. Its Theorem 1.3 gives $0.6$ for $0<\lambda<1$, with a constant depending on $\lambda$. At $\lambda=1$: a remark at the end of [1, Section 4] sketches $8/13=0.6153\ldots$; Li [7] obtains $56/85=0.6588\ldots$ from the method of [3], its limit without further optimisation by his account.

Theorem 1.

$N_{1}(X)\ll_{\varepsilon}X^{7/12+\varepsilon}$, where $7/12=0.5833\ldots$

Since $N_{\mu}\leq N_{1}$ for $\mu\leq 1$, Theorem 1 also improves [2, Theorem 1.3] on $0<\mu<1$, with a constant independent of $\mu$. It is the best known exponent for $61/69<\mu\leq 1$; for $\mu\leq 61/69$, (1.2) is at least as good.

Section 2 recalls the anatomic reduction of [3, 1, 2]. Fix $\varepsilon_{0}>0$ and $d=d(\varepsilon_{0})$. Every triple lies in a box. In a box, each side is a fixed coefficient times $\prod_{i\leq d}\mathfrak{s}_{i}^{\,i}$. The roots $\mathfrak{s}_{i}$ have fixed dyadic sizes. Distinct side values are coprime. Let $\mathcal{B}_{d}$ be the number of triples in a box and $D=\log_{X}\mathcal{B}_{d}$. The proof is the chain

$$\begin{gathered}N_{1}(X)\ll(\log X)^{2}\max_{Y\leq X,\ \mu\leq 1}N^{*}_{\mu}(Y),\qquad N^{*}_{\mu}(Y)\ll Y^{\varepsilon_{0}}\max\mathcal{B}_{d},\\ D\leq\frac{7}{12}+\frac{18}{7}\rho,\qquad\rho\leq 19\varepsilon_{0},\end{gathered}$$

from (2.1), (2.3), Theorem 2 with Lemma 7.1, and Lemma 7.2. Since $\tfrac{18}{7}\cdot 19+1=\tfrac{349}{7}$, choosing $\varepsilon_{0}$ with $\tfrac{349}{7}\varepsilon_{0}<\varepsilon/2$ proves Theorem 1. The inequalities for $D$ are linear in the $25$ exponents

(1.3) $$(a,b,c;\,a_{1},\dots,a_{6},b_{1},\dots,b_{6},c_{1},\dots,c_{6};\,v_{a},v_{b},v_{c};\,D),$$

with $L=a+b+c$. The $22$ inequalities of [2, Proposition 6.1] give $D\leq 3/5$ there.

We add two kinds of rows. First, the tail (Section 3). The root classes of index at least seven cost $X^{\tau_{s}}$ to enumerate, $\tau_{s}=s-\sum_{i\leq 6}s_{i}$. Their product is at least $X^{v_{s}}$, with $v_{s}\geq 7\tau_{s}$. So the tail gives two exact rows per side and can serve as a modulus. Second, the cut (Section 5). Choose a side $\sigma$, and let $y,z$ be the other two. Choose a nonempty $T\subseteq\{1,\ldots,6,\mathrm{tail}\}$ of root classes of $\sigma$, and $R_{y},R_{z}\subseteq[6]$, not both empty. The tail is never free. Enumerate all other roots. The selected roots form a modulus $M$. Put $q=\gcd(\gcd R_{y},\gcd R_{z})$, $\gcd\emptyset=0$, and

$$u=\prod_{r\in R_{y}}\mathfrak{y}_{r}^{\,r/q},\qquad v=\prod_{s\in R_{z}}\mathfrak{z}_{s}^{\,s/q},$$

with $\mathfrak{y}_{r},\mathfrak{z}_{s}$ the free roots of $y,z$; an empty product is $1$. Then $u\equiv\xi v\pmod{M}$ for one of $X^{o(1)}$ units $\xi$, and

$$\gcd(u,v)=1,\qquad u\equiv\xi v\pmod{M},\qquad 0<u\leq U,\quad 0<v\leq V,$$

a primitive point of a lattice of determinant $M$; Lemma 4.1 counts these by $1+2UV/M$. This is the cut (5.3). There is one per $(\sigma,T,R_{y},R_{z})$:

(1.4) $$3\,(64^{2}-1)\,127=1\,560\,195$$

cuts.

Section 6 proves $D\leq 7/12$ from these rows. The thirteen min-disjunctions of [2, Proposition 6.1] give $8192$ cells. Each cell closes along a binary tree of sign covers. Its leaves carry nonnegative rational Farkas multipliers. Only $424$ cells need a cut; $732$ distinct cuts occur. On a leaf the multipliers sum to at most $18/7$, whence the term $\tfrac{18}{7}\rho$. Section 7 uses it at $\lambda=1$. There the row $L\leq 1$ of [2] fails. Its proof in [2, Theorem 1.3] needs $\varepsilon_{0}\leq(1-\lambda)/3$. So for $\lambda<1$ the constants, $d$ among them, depend on $\lambda$; at $\lambda=1$ no $\varepsilon_{0}>0$ serves. We use $L\leq 1+3\varepsilon_{0}/4$ instead. The $L$-native Fourier and geometry rows cost $18\varepsilon_{0}$ and $6\varepsilon_{0}$. Hence $\rho=18\varepsilon_{0}+o_{\varepsilon_{0},d}(1)$.

The proof in [2] does not use the determinant method; [3] used it together with Thue equations. Lemma 4.1 is its simplest case. The cut applies it to two sides at once, with modulus $M$ built from roots on the third side.

Imported statements are cited where used. Citations of [2] refer to arXiv:2410.12234v2 (to appear in C. R. Math.); each imported statement was checked against it. The dyadic reduction (2.1) is only stated in [2, Section 2]; Section 2 proves it. Besides the divisor bound and its consequence for $x^{j}-y^{j}=n$ [2, Propositions 2.3 and 3.1], only [2, Proposition 4.1] rests on an outside result, [5, Lemma 3]. Nothing else in [2, Sections 2–6] is cited from elsewhere. No conjecture is assumed, and no estimate for rational points on curves is used. The finite step, Theorem 2, is a certificate checked by four independent programs (Appendix B). Every other step is proved in the text.

2. The anatomic reduction

Throughout this section $0<\lambda\leq 2$, $\varepsilon>0$ is arbitrary, $\varepsilon_{0}>0$ is small and fixed before $X$, and $d=d(\varepsilon_{0})$. Implied constants may depend on $\lambda,\varepsilon,\varepsilon_{0},d$, never on the box, the coefficients or $X$.

Write $N^{*}_{\lambda}(X)$ for the number of triples counted by $N_{\lambda}(X)$ that satisfy in addition $c\geq X/2$ and $\operatorname{rad}(abc)\geq c^{\lambda}/2$. A dyadic decomposition and the pigeonhole principle give

(2.1) $$N_{\lambda}(X)\ll_{\lambda}(\log X)^{2}\max_{1\leq Y\leq X}\ \max_{\mu\leq\lambda}N^{*}_{\mu}(Y),$$

as in [2, Section 2]. The bounds below for $N^{*}_{\mu}(Y)$ are uniform in $0<\mu\leq\lambda$ (Lemma 7.2).

Proof of (2.1).

Let $(a,b,c)$ be counted by $N_{\lambda}(X)$. Then $c\geq 2$ and $a,b<c\leq X$. Choose $l\geq 1$ with $c\in(X2^{-l},X2^{1-l}]$ and put $Y=X2^{1-l}\leq X$. Thus $Y/2<c\leq Y$. Put

$$\mu_{i}=\lambda-i/\log_{2}X\qquad(0\leq i\leq\lceil\lambda\log_{2}X\rceil).$$

Here $\log_{2}X=(\log X)/(\log 2)$. There are $O(\log X)$ such exponents, all at most $\lambda$, and the last is at most $0$. With $r=\operatorname{rad}(abc)$, we have $r<c^{\mu_{0}}$ and $r\geq 1\geq c^{\mu_{m}}$ for the last index $m$. Choose the largest $i$ with $r<c^{\mu_{i}}$. Then $r\geq c^{\mu_{i+1}}\geq c^{\mu_{i}}/2$, since $c\leq X$. Hence the triple is counted by $N^{*}_{\mu_{i}}(Y)$. There are $O(\log X)$ choices of $Y$ and $O(\log X)$ choices of $\mu_{i}$, which proves the claim. ∎

For $n\geq 1$ write $[n]=\{1,\ldots,n\}$.

Let $d\in\mathbb{N}$ and $(a_{0},b_{0},c_{0})\in\mathbb{N}^{3}$. Let $\mathbf{A},\mathbf{B},\mathbf{C}\in\mathbb{N}^{d}$ have entries powers of two, $1$ included. Put $\mathfrak{a}_{0}=a_{0}$, $\mathfrak{b}_{0}=b_{0}$, $\mathfrak{c}_{0}=c_{0}$ and

(2.2) $$\mathcal{B}_{d}=\#\Bigl{\{}(\mathfrak{a}_{i},\mathfrak{b}_{i},\mathfrak{c}_{i})_{i=1}^{d}\in\mathbb{N}^{3d}:\begin{aligned} &\mathfrak{a}_{i}\sim A_{i},\ \mathfrak{b}_{i}\sim B_{i},\ \mathfrak{c}_{i}\sim C_{i}\quad(1\leq i\leq d),\\ &a_{0}\textstyle\prod_{i\leq d}\mathfrak{a}_{i}^{\,i}+b_{0}\prod_{i\leq d}\mathfrak{b}_{i}^{\,i}=c_{0}\prod_{i\leq d}\mathfrak{c}_{i}^{\,i},\\ &\gcd\bigl{(}\textstyle\prod_{i=0}^{d}\mathfrak{a}_{i},\prod_{i=0}^{d}\mathfrak{b}_{i},\prod_{i=0}^{d}\mathfrak{c}_{i}\bigr{)}=1\end{aligned}\Bigr{\}}.$$

For a side write $S_{j}\in\{A_{j},B_{j},C_{j}\}$ for its dyadic root size. Equation (2.2) is the $B_{d}$ of [2], under our dyadic convention, with no condition within a side. We write $\mathcal{B}_{d}$ to avoid a clash with the size $B_{d}$.

Here $x\sim A$ means $A\leq x<2A$, as in [1]; [2] uses $x\in(A,2A]$. Both give $x/2\leq A\leq x$ and $\#\{x:x\sim A\}=A$, so no inequality of [2] changes. The difference is at $x=1$, which no interval $(A,2A]$, $A\in\mathbb{N}$, contains. Roots equal to $1$ occur: in [2, Lemma 2.2], $m_{j}=\prod_{p^{j}\|n}p$ and $n_{j}=m_{j}$ for $j\neq k$, the folded index. So $n_{j}=1$ unless some prime has multiplicity exactly $j$. As $n$ has at most $\omega(n)$ distinct multiplicities, all but at most $\omega(n)$ indices $j\leq d$, $j\neq k$, give $n_{j}=1$. So in the proof of [2, Proposition 2.1] we read “the unique powers of two such that $a_{i}\sim A_{i}$” with $A_{i}=1$ when $a_{i}=1$. By [2, Proposition 2.1], each $\varepsilon_{0}>0$ has a $d=d(\varepsilon_{0})$ with

(2.3) $$N^{*}_{\lambda}(X)\ll_{\varepsilon_{0}}X^{\varepsilon_{0}}\max\mathcal{B}_{d},$$

the maximum over pairwise coprime $1\leq a_{0},b_{0},c_{0}\leq X^{\varepsilon_{0}}$ and dyadic $A_{i},B_{i},C_{i}\in[1,X]$ with

(2.4) $$\begin{gathered}2^{-O_{d}(1)}X^{\lambda}\leq\prod_{j\leq d}A_{j}B_{j}C_{j}\leq X^{\lambda+3\varepsilon_{0}/4},\\ \prod_{j\leq d}A_{j}^{\,j}\leq X,\qquad\prod_{j\leq d}B_{j}^{\,j}\leq X,\qquad X^{1-\varepsilon_{0}/2}\leq\prod_{j\leq d}C_{j}^{\,j}\leq X,\end{gathered}$$

where the power of two depends only on $d$: by $\prod_{j}n_{j}\geq\operatorname{rad}(n)$ below and $c\geq X/2$, $\prod_{j}A_{j}B_{j}C_{j}\geq 2^{-3d}\operatorname{rad}(abc)\geq 2^{-3d-\lambda-1}X^{\lambda}$. This suffices, since $d$ is fixed before $X$. Fix such a box. For a side $s\in\{a,b,c\}$ set

(2.5) $$s_{i}=\log_{X}S_{i}\ (1\leq i\leq d),\qquad s=\log_{X}\!\prod_{i\leq d}S_{i}=\sum_{i\leq d}\log_{X}S_{i}.$$

Index $0$ is not a coordinate: $a_{0},b_{0},c_{0}$ are the integer coefficients. From here on $a,b,c$ denote coordinates, not the triple. Boxes with $\mathcal{B}_{d}=0$ do not attain the maximum in (2.3) unless $N^{*}_{\lambda}(X)=0$. So $\mathcal{B}_{d}\geq 1$, and we set

$$D=\log_{X}\mathcal{B}_{d}\geq 0.$$

For the tail put

(2.6) $$\tau_{s}=s-\sum_{i\leq 6}s_{i}=\sum_{7\leq j\leq d}\log_{X}S_{j},\qquad v_{s}=\sum_{7\leq j\leq d}j\log_{X}S_{j}.$$

Put $L=a+b+c$ and $p_{i}=a_{i}+b_{i}+c_{i}$; a bare $p$ is a prime. Moving a dyadic endpoint at scale $X$ changes the root coordinates by

(2.7) $$\Delta s_{i},\ \Delta s=O_{d}\!\left(\frac{1}{\log X}\right).$$

This does not cover $D$. Its resolution loss is in Lemma 7.2 and Table 7.1.

The left inequality of (2.4) uses $\prod_{j\leq d}n_{j}\geq\operatorname{rad}(n)$ with no loss, while [2, Lemma 2.2] states only $\operatorname{rad}(n)\leq n^{\varepsilon_{0}}\prod_{j}n_{j}$. In the proof of that lemma $m_{j}=\prod_{p^{j}\|n}p$, and, for the folded index $k$, $n_{j}=m_{j}$ for $j\neq k$ and $n_{k}=m_{k}\prod_{j>d}m_{j}^{\lfloor j/k\rfloor}$. Since $d\geq k$, $\lfloor j/k\rfloor\geq 1$ for $j>d$, so

$$\prod_{j\leq d}n_{j}\ \geq\ \prod_{j\leq d}m_{j}\prod_{j>d}m_{j}\ =\ \operatorname{rad}(n).$$

Lemma 2.1.

The three side values of a box are pairwise coprime. Hence every factor of one side is a unit modulo every factor of another side.

Proof.

A prime dividing two side values divides the third by the equation in (2.2). If it divides $a_{0}\prod_{i=1}^{d}\mathfrak{a}_{i}^{\,i}$, it divides $\prod_{i=0}^{d}\mathfrak{a}_{i}$; likewise on the other sides. This contradicts the gcd condition in (2.2). ∎

The index $i=0$ in this gcd condition is needed. With $d=1$, the choice

$$a_{0}=b_{0}=1,\quad\mathfrak{a}_{1}=\mathfrak{b}_{1}=2,\quad c_{0}=4,\quad\mathfrak{c}_{1}=1$$

satisfies the condition with $i\geq 1$: $\gcd(2,2,1)=1$. Yet its side values $2$, $2$, $4$ share the prime $2$. The full condition excludes it: $\gcd(2,2,4)=2$.

Nothing is assumed within a side. Roots of one side may share primes with $a_{0}$. Unenumerated roots are recovered by a divisor bound, not by squarefreeness.

2.1. The inherited rows

In the coordinates (2.5), the propositions of [2] give the $22$ inequalities of [2, Proposition 6.1]. We display them in exact limiting form, as Section 6 uses them. At finite resolution each right-hand side rises by the amount in Lemma 7.2 and Table 7.1; (2.9) and (2.10) hold exactly.

The radical bound in (2.4) gives [2, (6.1)],

(2.8) $$L\leq 1,$$

derived there for $\lambda<1$. At $\lambda=1$ only $L\leq 1+3\varepsilon_{0}/4$ holds (Lemma 7.2). Roots sum into their side, [2, (6.2)]:

(2.9) $$\sum_{i\leq 6}s_{i}\leq s\qquad(s=a,b,c).$$

The side-value bound in (2.4) gives [2, (6.3)],

(2.10) $$7s-1\leq 6s_{1}+5s_{2}+4s_{3}+3s_{4}+2s_{5}+s_{6}\qquad(s=a,b,c).$$

Two sides determine the third up to a divisor factor [2, Proposition 2.3]; this is [2, (6.4)],

(2.11) $$D\leq\min(a+b,\ a+c,\ b+c).$$

Geometry of numbers [2, Proposition 4.1] gives the thirteen min-disjunctions [2, (6.5)–(6.9)], labelled (G1)–(G13) in Appendix A. The Fourier estimate [2, Proposition 3.1] gives [2, (6.10)],

(2.12) $$\sum_{s=a,b,c}\max(s_{2}+s_{4}+s_{6},\ s_{3}+s_{6},\ s_{5})\leq 4L-6D.$$

One argument per maximum gives $27$ linear rows; Section 6 uses these.

The Python listing of [2, Appendix A] transcribes the first line of [2, (6.9)] as p1+a2+b3+a3, where the displayed formula reads $p_{1}+a_{2}+b_{3}+c_{3}$. We use the displayed formula. It is what [2, Proposition 4.1] gives with the printed second argument $1-a_{2}-2b_{3}-2c_{3}$. For (2.10) we use

(2.13) $$7s-1\leq 7s-\sum_{i\leq d}i\,s_{i}=\sum_{i\leq 6}(7-i)s_{i}+\sum_{7\leq i\leq d}(7-i)s_{i}\leq\sum_{i\leq 6}(7-i)s_{i},$$

since $7-i\leq 0$ for $i\geq 7$. In our system (2.10) is the sum of the two tail rows of Section 3, so it is redundant.

3. The multiplicity-at-least-seven tail

The tail rows we use are

(3.1) $$\tau_{s}\geq 0,\qquad v_{s}\geq 7\tau_{s},\qquad\sum_{i\leq 6}i\,s_{i}+v_{s}\leq 1\qquad(s=a,b,c).$$

The first is (2.9). The second: $\log_{X}S_{j}\geq 0$, with weight $j\geq 7$ in $v_{s}$ and weight $1$ in $\tau_{s}$. The third is the side-value bound in (2.4). Adding the last two,

$$(v_{s}-7\tau_{s})+(1-\sum_{i\leq 6}is_{i}-v_{s})\geq 0,$$

that is, $\sum_{i\leq 6}is_{i}+7\tau_{s}\leq 1$; with $\tau_{s}=s-\sum_{i\leq 6}s_{i}$ this is (2.10). With the upper row $v_{s}\leq d\tau_{s}$, these rows describe the real pairs $(\tau_{s},v_{s})$ exactly. We omit that row, so the system $\mathcal{S}$ of Section 6 is free of $d$.

Lemma 3.1.

Fix a side $s\in\{a,b,c\}$ and let $M_{\mathrm{tail}}=\prod_{7\leq j\leq d}\mathfrak{s}_{j}^{\,j}$ with $\mathfrak{s}_{j}\sim S_{j}$. Then

(3.2) $$\begin{gathered}\#\{(\mathfrak{s}_{7},\dots,\mathfrak{s}_{d})\in\mathbb{N}^{d-6}:\mathfrak{s}_{j}\sim S_{j}\}\leq\prod_{7\leq j\leq d}S_{j}=X^{\tau_{s}},\\ M_{\mathrm{tail}}\geq\prod_{7\leq j\leq d}S_{j}^{\,j}=X^{v_{s}},\end{gathered}$$

and every prime $p$ dividing $M_{\mathrm{tail}}$ does so with $\operatorname{ord}_{p}(M_{\mathrm{tail}})\geq 7$, where $\operatorname{ord}_{p}$ denotes the $p$-adic valuation.

Proof.

Each interval $[S_{j},2S_{j})$ contains exactly $S_{j}$ integers, which gives the count, and $\mathfrak{s}_{j}\geq S_{j}$ gives the modulus bound. If $p\mid\mathfrak{s}_{j}$ with $j\geq 7$ then $\operatorname{ord}_{p}(M_{\mathrm{tail}})\geq j\geq 7$. ∎

The count is exact for $[N,2N)$ and for $(N,2N]$. A closed $[N,2N]$ would overcount by a factor of up to $2^{d-6}$, attained when all $S_{j}=1$.

So the tail costs $X^{\tau_{s}}$ to enumerate. The bound is on roots, not on $\operatorname{rad}(M_{\mathrm{tail}})$, and uses no squarefreeness. On the modulus side $\sigma$ of Section 5, $M_{\mathrm{tail}}$ divides the side value and is at least $X^{v_{\sigma}}$. Hence $T$ may contain the tail.

3.1. Folded roots

In [2, Lemma 2.2] the very high multiplicities are folded into one index $k$. The folded root need not be squarefree; the remainder goes into $n_{0}$. The proof of [2, Proposition 2.1] uses that lemma with parameter $\varepsilon_{0}/4$. So

$$k=\lceil 4/\varepsilon_{0}\rceil,\qquad d=6\lceil(4/\varepsilon_{0})^{2}\rceil.$$

We may assume $\varepsilon_{0}\leq 4/7$. Then $k\geq 7$, and $d\geq 6(4/\varepsilon_{0})^{2}>4/\varepsilon_{0}+1>k$. So the folded root lies in the tail, $M_{\mathrm{tail}}$ still divides the side value, and $\operatorname{ord}_{p}(n_{k}^{\,k})=k\operatorname{ord}_{p}(n_{k})\geq 7$ for $p\mid n_{k}$. The overlap of $n_{0}$ and $n_{k}$ is within one side, so Lemma 2.1 is untouched. Unenumerated roots are recovered by Lemma 4.3. The folding costs two things, both already charged: the factor $X^{3\varepsilon_{0}/4}$ in the upper radical bound of (2.4), and the choice of $a_{0},b_{0},c_{0}$ in the $X^{\varepsilon_{0}}$ of (2.3). It changes neither Lemma 3.1 nor (3.1).

4. Three lattice and divisor lemmas

Lemma 4.1.

Let $M\geq 1$, let $\xi$ be a unit modulo $M$ and let $U,V\geq 1$. Then

(4.1) $$\#\{(u,v)\in\mathbb{N}^{2}:u\leq U,\ v\leq V,\ \gcd(u,v)=1,\ u\equiv\xi v\ (M)\}\leq 1+\frac{2UV}{M}.$$

Proof.

Put $\mathcal{L}_{\xi}=\{(u,v)\in\mathbb{Z}^{2}:u\equiv\xi v\ (M)\}$, of index $M$ in $\mathbb{Z}^{2}$. Any two of its points have determinant in $M\mathbb{Z}$. Positive collinear primitive points coincide. So the primitive points of $\mathcal{L}_{\xi}$ in $[1,U]\times[1,V]$ have distinct slopes; order them by slope. Consecutive points span with $0$ a triangle of area $\geq M/2$. These triangles have disjoint interiors and lie in $[0,U]\times[0,V]$. With $P$ points, $(P-1)M/2\leq UV$. ∎

The proof does not use that $\xi$ is a unit. The constant $2$ is sharp. For odd $n\geq 3$ take $U=n$, $V=n+4$, $M=n^{2}+2n-4$, $\xi=n-1$. The primitive points are $(n-1,1)$, $(n,n+4)$, $(1,n+3)$, with consecutive determinants $M$, and $(P-1)M/(UV)\to 2$.

Lemma 4.2.

Let $1\leq q\leq 6$, let $M\geq 1$ and let $K$ be a unit modulo $M$. The congruence $\eta^{q}\equiv K\pmod{M}$ has at most $8^{\omega(M)}$ solutions in units $\eta$, hence $X^{o(1)}$ solutions uniformly for $M\leq X^{O(1)}$.

Proof.

Odd prime power: the unit group is cyclic, so the kernel of $\eta\mapsto\eta^{q}$ has order at most $q\leq 6$. Modulus $2^{e}$, $e\leq 2$: the unit group has order at most $2$. Modulus $2^{e}$, $e\geq 3$: $(\mathbb{Z}/2^{e}\mathbb{Z})^{\times}\cong C_{2}\times C_{2^{e-2}}$, so the kernel has order $\gcd(q,2)\gcd(q,2^{e-2})\leq 8$; the $2$-part of $q\leq 6$ is at most $4$. Multiply by the Chinese remainder theorem. As $\omega(M)\ll\log M/\log\log M$ for $M\geq 3$,

$$8^{\omega(M)}=\exp\bigl{(}O(\log M/\log\log M)\bigr{)}=X^{O(1/\log\log X)}\qquad(M\leq X^{O(1)}).$$

The form $C^{\omega(M)}\ll_{C,\delta}M^{\delta}$, for fixed $C\geq 1$ and $\delta>0$, alone gives no rate. For $M\leq 2$ there is one class. No bound $B^{\omega(M)}$ with $B<8$ holds. For $q=4$, $e\geq 4$, the group $C_{2}\times C_{2^{e-2}}$ has $\gcd(4,2)\gcd(4,2^{e-2})=8$ fourth roots of unity. And $q=4$ occurs in (5.2). Local kernel size $8$ occurs only for $2^{e}$, $e\geq 4$, $q=4$. So for $\omega(M)\geq 2$ the product of local kernel sizes is at most $8\cdot 6^{\omega(M)-1}<8^{\omega(M)}$. ∎

Lemma 4.3.

Write $\operatorname{div}(N)=\#\{m\in\mathbb{N}:m\mid N\}$ for $N\in\mathbb{N}$. For $r\geq 1$ and positive integers $e_{1},\dots,e_{r}$,

$$\#\{(x_{1},\dots,x_{r})\in\mathbb{N}^{r}:x_{1}^{e_{1}}\cdots x_{r}^{e_{r}}=N\}\leq\operatorname{div}(N)^{r-1}\ll_{r,\delta}N^{\delta}\qquad(\delta>0),$$

uniformly in $e_{1},\dots,e_{r}$. For $N\geq 3$ the maximal-order bound gives the rate

$$\operatorname{div}(N)^{r-1}=\exp\!\bigl{(}O_{r}(\log N/\log\log N)\bigr{)},$$

so for $N\leq X^{O_{d}(1)}$ and $r\leq d$ the count is $X^{O_{d}(1/\log\log X)}$.

Proof.

Each $x_{i}$ divides $N$. The equation determines $x_{r}$ from $x_{1},\dots,x_{r-1}$, for any $e_{i}$. Apply the divisor bound and, for the rate, the maximal order of $\operatorname{div}(N)$. ∎

Thus an unfactored product of roots determines those roots up to $X^{o(1)}$.

5. The cut

Fix a box and $(\sigma,T,R_{y},R_{z})$. Let $(y,z)$ be the other two sides in alphabetical order: $(b,c)$, $(a,c)$, $(a,b)$ for $\sigma=a,b,c$. Write $[6]=\{1,2,3,4,5,6\}$. Choose a nonempty

(5.1) $$\begin{gathered}T\subseteq\{1,2,3,4,5,6,\mathrm{tail}\},\\ C_{T}=\sum_{t\in T\cap[6]}\sigma_{t}+\mathbf{1}_{\mathrm{tail}\in T}\tau_{\sigma},\qquad W_{T}=\sum_{t\in T\cap[6]}t\,\sigma_{t}+\mathbf{1}_{\mathrm{tail}\in T}v_{\sigma}.\end{gathered}$$

Choose free sets $R_{y},R_{z}\subset[6]$, not both empty, and put

(5.2) $$q=\gcd(\gcd R_{y},\gcd R_{z}),\qquad F=\sum_{r\in R_{y}}r\,y_{r}+\sum_{s\in R_{z}}s\,z_{s},$$

with $\gcd\emptyset=0$. As $R_{y}\cup R_{z}\neq\emptyset$, $1\leq q\leq 6$ and $q$ divides every element of $R_{y}\cup R_{z}$. By Section 3.1, $d=6\lceil(4/\varepsilon_{0})^{2}\rceil\geq 294$, so the tail notation applies. Write $\max\{t,0\}=\max\{t,0\}$.

Proposition 5.1.

Let a box satisfy the constraints after (2.3), with $\mathcal{B}_{d}\geq 1$. Then

(5.3) $$D\leq\Bigl{(}y-\sum_{r\in R_{y}}y_{r}\Bigr{)}+\Bigl{(}z-\sum_{s\in R_{z}}z_{s}\Bigr{)}+C_{T}+\max\{\frac{F}{q}-W_{T},0\}+o_{\varepsilon_{0},d}(1),$$

uniformly over the finitely many choices of $(\sigma,T,R_{y},R_{z})$.

For later use write

$$\mathsf{A}=\Bigl{(}y-\sum_{r\in R_{y}}y_{r}\Bigr{)}+\Bigl{(}z-\sum_{s\in R_{z}}z_{s}\Bigr{)}+C_{T},\qquad\mathsf{B}=F/q-W_{T}.$$

The limiting form of (5.3) is $D\leq\mathsf{A}+\max(0,\mathsf{B})$.

Proof.

Sizes. Let $\sigma_{0},y_{0},z_{0}$ be the coefficients and $\Sigma_{i},Y_{i},Z_{i}$ the dyadic root sizes of the $\sigma,y,z$ sides. Besides (2.2), the proof uses the dyadic ranges and $\prod_{i}A_{i}^{i}$, $\prod_{i}B_{i}^{i}$, $\prod_{i}C_{i}^{i}\leq X$. So all products and cofactors are at most $X^{O_{d}(1)}$, and Lemmas 4.2 and 4.3 cost $X^{o(1)}$. The coefficients enter no size estimate.

Enumeration. Write $m_{i},n_{i},n^{\prime}_{i}$ for the roots of the $\sigma$-, $y$- and $z$-sides, so $m_{i}\sim\Sigma_{i}$, $n_{i}\sim Y_{i}$, $n^{\prime}_{i}\sim Z_{i}$. The coefficients $a_{0},b_{0},c_{0}$ are fixed by the box. Enumerate: the roots of $y$ outside $R_{y}$; of $z$ outside $R_{z}$; the selected roots of $\sigma$. Tails go through Lemma 3.1. The $y$- and $z$-tails are never in $R_{y},R_{z}\subseteq[6]$, so they are enumerated too. The cost is

(5.4) $$X^{(y-\sum_{r\in R_{y}}y_{r})+(z-\sum_{s\in R_{z}}z_{s})+C_{T}+o_{\varepsilon_{0},d}(1)}.$$

Modulus. The tail roots of $\sigma$ are $m_{j}$, $j\geq 7$. Put

(5.5) $$M=\prod_{t\in T\cap[6]}m_{t}^{\,t}\,M_{\mathrm{tail}}^{\mathbf{1}_{\mathrm{tail}\in T}},\qquad M_{\mathrm{tail}}=\prod_{7\leq j\leq d}m_{j}^{\,j}.$$

Then $M$ divides the $\sigma$-side value. Also $m_{t}\geq\Sigma_{t}=X^{\sigma_{t}}$ (left-closed intervals), and Lemma 3.1 bounds the tail. So

(5.6) $$M\geq X^{W_{T}}.$$

Put

(5.7) $$u=\prod_{r\in R_{y}}n_{r}^{\,r/q},\qquad v=\prod_{s\in R_{z}}(n^{\prime}_{s})^{\,s/q},$$

an empty product being $1$; the exponents are integers by (5.2).

Congruence. By Lemma 2.1, modulo $M$ every factor of the $y$- and $z$-sides is a unit: coefficients, enumerated roots, $u$, $v$. The equation of (2.2) modulo $M$ gives

(5.8) $$(uv^{-1})^{q}\equiv K\pmod{M}$$

with

(5.9) $$K\equiv\epsilon E_{z}E_{y}^{-1}\pmod{M},\qquad\epsilon=\begin{cases}-1,&\sigma=c,\\ +1,&\sigma\neq c,\end{cases}$$

where

$$E_{y} =y_{0}\prod_{i\in[6]\setminus R_{y}}n_{i}^{\,i}\prod_{7\leq j\leq d}n_{j}^{\,j},$$

$$E_{z} =z_{0}\prod_{i\in[6]\setminus R_{z}}(n^{\prime}_{i})^{\,i}\prod_{7\leq j\leq d}(n^{\prime}_{j})^{\,j}.$$

So $K$ is a unit fixed by the enumerated data.

Count. By Lemma 4.2, $\xi=uv^{-1}$ lies in one of $X^{o(1)}$ classes. By Lemma 2.1, $\gcd(u,v)=1$. Also

$$u\leq 2^{\sum_{r\in R_{y}}r/q}\prod_{r\in R_{y}}Y_{r}^{r/q}=:U,\qquad v\leq 2^{\sum_{s\in R_{z}}s/q}\prod_{s\in R_{z}}Z_{s}^{s/q}=:V,$$

with $U,V\geq 1$; the powers of $2$ cost $O_{d}(1/\log X)$, as $\sum_{r\leq 6}r=21$. Hence

$$U=X^{\sum_{r\in R_{y}}r\,y_{r}/q+O_{d}(1/\log X)},\qquad V=X^{\sum_{s\in R_{z}}s\,z_{s}/q+O_{d}(1/\log X)},$$

so $UV=X^{F/q+O_{d}(1/\log X)}$, and Lemma 4.1 with (5.6) bounds the pairs $(u,v)$ in one class by

(5.10) $$1+\frac{2UV}{M}\ll_{d}X^{\max\{F/q-W_{T},0\}+O_{d}(1/\log X)}.$$

Recovery. Given $u,v$, Lemma 4.3 recovers the free roots up to $X^{o(1)}$. Then the $y$- and $z$-side values are known, and the equation fixes the $\sigma$-side value. Divide by $\sigma_{0}$ and the selected roots:

$$N=\prod_{i\in[6]\setminus T}m_{i}^{\,i}\,\left(\prod_{7\leq j\leq d}m_{j}^{\,j}\right)^{1-\mathbf{1}_{\mathrm{tail}\in T}}.$$

Lemma 4.3 recovers these unselected roots up to $X^{o(1)}$. Multiplying (5.4) by $X^{o(1)}$ and (5.10) gives (5.3). ∎

If one free set is empty, $U=1$ or $V=1$, and Lemma 4.1 still applies. The exponent $q$ depends only on the index sets $R_{y},R_{z}$, not on which roots equal $1$. Counting coefficients is charged to (2.3): inside a box $a_{0},b_{0},c_{0}$ are fixed.

Count: $64$ subsets of $[6]$ per counted side, $127$ sets $T$, three sides $\sigma$. Without the pair of empty free sets this is $3(64^{2}-1)127=1\,560\,195$, as in (1.4).

6. The finite system

Let $\mathcal{S}$ be the following system in the $25$ nonnegative variables (1.3):

• (2.8), (2.9), (2.10), (2.11), the $27$ rows of (2.12), and the last two rows of (3.1) per side: $43$ linear rows (the first row of (3.1) is (2.9));

• the thirteen min-disjunctions (G1)–(G13) of Appendix A;

• the $1\,560\,195$ cuts (5.3), without the $o_{\varepsilon_{0},d}(1)$.

Here $L$, $p_{i}$ and $\tau_{s}$ are linear forms, not coordinates. The coefficients $a_{0},b_{0},c_{0}$ are not coordinates either. Coefficients, dyadic product bounds, tail enumeration and the $X^{o(1)}$ recovery costs are all inside these rows. So the theorem below is a finite statement about (1.3).

Theorem 2.

Every nonnegative solution of $\mathcal{S}$ satisfies $D\leq 7/12$.

Proof.

Write $x$ for the $25$-vector in (1.3), and write $e_{D}$ for the unit vector in its $D$ coordinate. Each min-disjunction holds in one of its two branches; branch $1$ of (G$j$) is $D\leq L-P_{j}$ and branch $2$ is $D\leq L-Q_{j}$. For $\eta\in\{1,2\}^{13}$ let the cell $\mathcal{C}_{\eta}$ be the intersection of the selected branches. A feasible point lies in at least one of these $8192$ cells; cells may overlap. In a cell, a cut $D\leq\mathsf{A}+\max(0,\mathsf{B})$ is covered exactly by

(6.1) $$\{D\leq\mathsf{A}+\mathsf{B}\}\ \cup\ \{\mathsf{B}\leq 0,\ D\leq\mathsf{A}\}.$$

For each cell the certificate gives a full binary tree. Completeness is recursive: either the sole path is empty, or all paths begin with one cut descriptor, both $+$ and $-$ children occur, and the same holds for both suffix sets. The checker enforces: distinct leaf paths; no leaf a prefix of another; no descriptor repeated on a path. Let $x\in\mathcal{S}$ lie in the cell. By (6.1), $x$ descends at each node into a child. So it reaches a leaf whose rows all hold at $x$. A $+$ edge gives $D\leq\mathsf{A}+\mathsf{B}$; a $-$ edge gives $\mathsf{B}\leq 0$ and $D\leq\mathsf{A}$. At $\mathsf{B}=0$ either edge is allowed.

At a leaf $\ell$ of $\mathcal{C}$, the set $I(\mathcal{C},\ell)$ consists of: the $43$ linear rows; the thirteen branches of $\mathcal{C}$; the rows its path authorizes. No other row is admissible. All rows of $I(\mathcal{C},\ell)$ hold at $x$. The certificate gives $\lambda_{i}\geq 0$ in $\mathbb{Q}$ on $I(\mathcal{C},\ell)$, and $\lambda_{i}=0$ elsewhere. Write each row as $\langle g_{i},x\rangle\leq h_{i}$, and put $V_{\ell}=\sum_{i\in I(\mathcal{C},\ell)}\lambda_{i}h_{i}$. The checker verifies at every leaf

(6.2) $$\sum_{i}\lambda_{i}g_{i}\geq e_{D}\ \text{componentwise},\qquad\sum_{i}\lambda_{i}h_{i}\leq\frac{7}{12}.$$

All variables being nonnegative,

(6.3) $$D=\langle e_{D},x\rangle\leq\Bigl{\langle}\sum_{i}\lambda_{i}g_{i},x\Bigr{\rangle}\leq\sum_{i}\lambda_{i}h_{i}\leq\frac{7}{12}.\qed$$

The $43$ linear rows alone give $D\leq\tfrac{2}{3}$ (bound each maximum in (2.12) below by $0$). Adding the disjunctions gives the exact optimum $3/5$, the $0.6$ of [2, Theorem 1.3]. Adding the cuts gives $7/12$. Gains: $1/15$ from the disjunctions, $1/60$ from the cut, $1/12$ in total.

A leaf by hand. Take branch 2 of (G2) and (G3), branch 1 of (G5), any branch of the others. Multipliers: $1$ on (2.8), $1/4$ on the $Q$ arms of (G2) and (G3), $1/2$ on the $P$ arm of (G5). Then

$$\sum_{i}\lambda_{i}g_{i}=e_{D}+\tfrac{1}{2}(e_{a_{1}}+e_{b_{1}}+e_{c_{1}})\geq e_{D},\qquad\sum_{i}\lambda_{i}h_{i}=\tfrac{1}{2}\leq\tfrac{7}{12},\qquad\sum_{i}\lambda_{i}=2\leq\tfrac{18}{7}.$$

The coefficients of $a,b,c,p_{2},a_{3},b_{3}$ cancel; $D$ has $1$ and $p_{1}$ has $1/2$. So this leaf gives $D\leq 1/2$. Three bits are fixed, so the vector is valid on $2^{10}=1024$ cells; the certificate uses it on $200$.

The certificate is cert_leaves_tail.jsonl (Appendix B). Rebuild every row from (2.8)–(2.12), (3.1), (5.3) and (G1)–(G13); check (6.2) exactly. This gives Table 6.1. The $o_{\varepsilon_{0},d}(1)$ dropped from (5.3) is paid in Lemma 7.2. Put

$$\Lambda:=\max_{\text{certificate leaves}}\sum_{i}\lambda_{i},$$

the sum over all multipliers at the leaf, sign rows $\mathsf{B}\leq 0$ included.

Table 6.1. The certificate of Theorem 2. A no-cut cell has a sole empty path; otherwise it has a nontrivial complete cut trie. The certificate value is $V_{\ell}$ from the proof.

branch cells $\eta\in\{1,2\}^{13}$	$8192$
cells with a sole empty path (no cut)	$7768$
cells with a nontrivial complete cut trie	$424$
leaves	$10\,820$
internal binary nodes	$2628$
$\Lambda$	$18/7$
distinct cuts (5.3) used	$732$
worst certificate value	$7/12$
leaves with certificate value exactly $7/12$	$89$
minimum over leaves and coordinates of $\bigl{(}\sum_{i}\lambda_{i}g_{i}-e_{D}\bigr{)}_{k}$	$0$

6.1. The optimum

The bound $7/12$ is the exact optimum of $\mathcal{S}$. For $0\leq t\leq 1/24$, let $x(t)$ be the point

(6.4) $$\begin{gathered}a=\tfrac{1}{3},\ a_{1}=\tfrac{1}{8},\ a_{4}=\tfrac{5}{24},\ v_{a}=t;\qquad b=\tfrac{1}{4},\ b_{1}=\tfrac{1}{8},\ v_{b}=\tfrac{7}{8};\\ c=\tfrac{5}{12},\ c_{1}=\tfrac{1}{8},\ c_{3}=\tfrac{7}{24};\qquad D=\tfrac{7}{12},\end{gathered}$$

with every unlisted coordinate zero. The restriction $t\leq 1/24$ is $a_{1}+4a_{4}+v_{a}=23/24+t\leq 1$. For each such $t$: $L=1$; the $43$ linear rows hold; one arm of each (G$j$) holds. All $1\,560\,195$ cuts hold, with minimum bound exactly $7/12$. Here $t=v_{a}$ enters (5.3) only via $W_{T}$ ($\sigma=a$, $\mathrm{tail}\in T$), where it lowers $\mathsf{B}$. So each cut bound is nonincreasing in $t$, and $t=0$, $t=1/24$ suffice. At $t=0$ the minimum is attained $170\,335$ times. The cut family is $b\leftrightarrow c$ symmetric, and $b\leftrightarrow c$ maps $x(1/24)$ to

(6.5) $$\begin{gathered}a=\tfrac{1}{3},\ a_{1}=\tfrac{1}{8},\ a_{4}=\tfrac{5}{24},\ v_{a}=\tfrac{1}{24};\qquad b=\tfrac{5}{12},\ b_{1}=\tfrac{1}{8},\ b_{3}=\tfrac{7}{24};\\ c=\tfrac{1}{4},\ c_{1}=\tfrac{1}{8},\ v_{c}=\tfrac{7}{8};\qquad D=\tfrac{7}{12},\end{gathered}$$

with every unlisted coordinate zero. There it is attained $227\,551$ times. The points $x(0)$ and (6.5) lie in different cells. Their arms differ in (G3), (G4), (G5), (G7), (G9), (G10), (G12), (G13). These are points of $\mathcal{S}$, not $abc$ triples; they give no lower bound for $N_{1}(X)$.

7. Stability and the endpoint

Lemma 7.1.

Let $\rho\geq 0$ and $x\geq 0$. Suppose that, with right-hand sides raised by $\rho$,

• every linear row of $\mathcal{S}$ holds at $x$;

• at least one arm of each min-disjunction holds at $x$;

• every cut holds at $x$ in the form $D\leq\mathsf{A}+\max(0,\mathsf{B})+\rho$.

The sign rows $\mathsf{B}\leq 0$ of the descent are case distinctions and are not relaxed. Then

(7.1) $$D\leq\frac{7}{12}+\frac{18}{7}\rho.$$

Proof.

Each relaxed min-disjunction has a valid arm; choose a relaxed cell. The relaxed sign cover descends to a leaf as in Theorem 2. Then (6.3) with relaxed right-hand sides adds

$$\rho\sum_{i\ {\rm non\text{-}sign}}\lambda_{i}\leq\rho\sum_{i}\lambda_{i}\leq\Lambda\rho.$$

Table 6.1 gives $\Lambda=18/7$. For a relaxed cut the cover (6.1) becomes $\{D\leq\mathsf{A}+\mathsf{B}+\rho\}\cup\{\mathsf{B}\leq 0,\ D\leq\mathsf{A}+\rho\}$. The sign row stays exact, so nothing is relaxed twice. ∎

Lemma 7.2.

Let $0<\mu\leq 1$. Choose $Y_{0}=Y_{0}(\varepsilon_{0},d)$ such that, for $Y\geq Y_{0}$, each displayed loss $O_{d}(1/\log Y)$, $O_{d}(1/\log\log Y)$, $o_{\varepsilon_{0},d}(1)$ is at most $\varepsilon_{0}$. Let $Y\geq Y_{0}$, and take a nonempty box admissible in (2.3) at scale $Y$. Then every row of $\mathcal{S}$ holds in a $\mu$-free form. Its right-hand side rises by at most the amount in Table 7.1. In particular (2.8) becomes $L\leq 1+3\varepsilon_{0}/4$. Consequently

(7.2) $$\rho=18\varepsilon_{0}+o_{\varepsilon_{0},d}(1)$$

is admissible in Lemma 7.1. Increasing $Y_{0}$ if necessary, $\rho\leq 19\varepsilon_{0}$ for $\mu\leq 1$ and $Y\in[Y_{0},X]$.

Table 7.1. The relaxation carried by each family of $\mathcal{S}$.

family	source	raised by
(2.8)	[2, (6.1)]	$3\varepsilon_{0}/4$
(2.9), (2.10), (3.1)	definitions and (2.4)	$0$
(2.11)	Lemma 4.3	$O_{d}(1/\log\log Y)$
Appendix A	[2, Proposition 4.1]	$6\varepsilon_{0}+o_{\varepsilon_{0},d}(1)$
(2.12)	[2, Proposition 3.1], $D$-coefficient $6$	$18\varepsilon_{0}+o_{\varepsilon_{0},d}(1)$
(5.3)	Proposition 5.1	$o_{\varepsilon_{0},d}(1)$

Proof.

Run Section 2 at scale $Y$ with $\lambda$ replaced by $\mu$ and $\log_{Y}$ in place of $\log_{X}$ in (2.5). In (2.4), $\mu$ occurs only in the radical constraint. Its upper half is

(7.3) $$L\leq\mu+3\varepsilon_{0}/4\leq 1+3\varepsilon_{0}/4,$$

the one row of $\mathcal{S}$ that depends on $\mu$. Its lower half is

(7.4) $$\mu\leq L+O_{d}\Bigl{(}\frac{1}{\log Y}\Bigr{)}.$$

The proof of [2, Theorem 1.3] uses the radical bound in the weaker form

$$\prod_{j\leq d}A_{j}B_{j}C_{j}\leq Y^{\mu+3\varepsilon_{0}}\leq Y.$$

This needs $\varepsilon_{0}\leq(1-\mu)/3$, which fails at $\mu=1$ for every $\varepsilon_{0}>0$. It is the only use of $\mu<1$ in the proof of [2, Theorem 1.3], and the source of its $\mu$-dependent constant. We keep (7.3) as a relaxed row instead.

By (7.4), $\mu$ leaves the other rows. The other constraints of (2.4) are $\mu$-free. So (2.9), (2.10), (3.1) and Lemma 3.1 are $\mu$-free and exact. The constants in [2, Propositions 2.1, 3.1, 4.1] depend only on $\varepsilon_{0}$ and $d$.

Pair row. [2, Proposition 2.3] has no rate in $Y$; we prove (2.11) directly. There are exactly $\prod_{i}A_{i}B_{i}=Y^{a+b}$ choices of the $a$- and $b$-roots. With $c_{0}$ fixed, the equation in (2.2) then gives $N=\prod_{i}\mathfrak{c}_{i}^{\,i}\leq 2^{O_{d}(1)}Y$, and Lemma 4.3 leaves $\operatorname{div}(N)^{d-1}=Y^{O_{d}(1/\log\log Y)}$ choices. Likewise for the other two pairs. So (2.11) costs $O_{d}(1/\log\log Y)$, as in Table 7.1.

Cut. Proposition 5.1 uses only the dyadic ranges, (2.2) and the fixed-$d$ lemmas. So it holds verbatim at scale $Y$ for every $\mu\leq 1$. Its divisor and root-class factors cost $O_{d}(1/\log\log Y)$.

Fourier and geometry. The $\varepsilon_{0}$-exponents are $3\varepsilon_{0}$ in [2, Proposition 3.1], and $4\varepsilon_{0}$, $6\varepsilon_{0}$ in [2, Proposition 4.1]. There $\mu$ appears as $Y^{2\mu/3}$, resp. $Y^{\mu}$ and $Y^{\mu-1}$. By (7.4), $\mu\to L$ costs $4\cdot O_{d}(1/\log Y)$, resp. $O_{d}(1/\log Y)$. Our Fourier row is $\theta$ times the source inequality, with

$$\theta=\frac{[D]_{\text{row}}}{[D]_{\text{source}}}=\frac{6}{1}=\frac{[L]}{[\mu]}=\frac{4}{2/3}=\frac{1}{1/6}=6$$

on the $D$, $L$ and root terms. So $3\varepsilon_{0}$ becomes $18\varepsilon_{0}$. The rows (G1)–(G13) have coefficient $1$ on $D$ and $L$, so they rise by at most $6\varepsilon_{0}$.

These are the entries of Table 7.1; the largest is (7.2). All constants depend only on $\varepsilon_{0}$ and $d$, not on $\mu$. The $Y$-dependent errors are the displayed $O_{d}(1/\log Y)$, $O_{d}(1/\log\log Y)$ and $o_{\varepsilon_{0},d}(1)$ terms. They include the constants of [2, Propositions 3.1, 4.1], as in Appendix A. $Y\geq Y_{0}$ absorbs them.

∎

Lemma 7.2 removes the restriction $\lambda<1$ of [2, Theorem 1.3]. At $\mu=1$ the row (2.8) is lost, and Lemma 7.1 prices the relaxation $\rho$ at $\tfrac{18}{7}\rho$ in the exponent. The constant does not depend on $\mu$. $\mathcal{S}$, its certificate and $18/7$ are free of $\varepsilon_{0}$ and $d$. So choosing $\varepsilon_{0}$ from $18/7$ is not circular.

8. Proof of Theorem 1

Fix $\varepsilon>0$. Choose $\varepsilon_{0}>0$ with $\tfrac{349}{7}\varepsilon_{0}<\varepsilon/2$ and with folded index at least seven (Section 3.1). This fixes $d=d(\varepsilon_{0})$. The factors $8^{\omega(M)}$ and $\operatorname{div}(N)^{O_{d}(1)}$ are $Y^{o(1)}$ for fixed $d$ (Lemmas 4.2, 4.3); they need no budget. Dyadic constants and the count of boxes are inside (2.3).

Let $\mu\leq 1$ and $Y\leq X$. If $\mu\leq 0$, then $N^{*}_{\mu}(Y)=0$, as $\operatorname{rad}(abc)\geq 1\geq c^{\mu}$. Let $0<\mu\leq 1$. If $Y<Y_{0}(\varepsilon_{0},d)$, then $N^{*}_{\mu}(Y)\leq Y^{3}\ll_{\varepsilon_{0},d}1$. If $Y\geq Y_{0}$, then for a nonempty box Lemma 7.2 raises every row of $\mathcal{S}$ by at most $\rho\leq 19\varepsilon_{0}$. So the stability and reduction losses are at most

$$\frac{18}{7}\,19\varepsilon_{0}+\varepsilon_{0}=\frac{349}{7}\varepsilon_{0}<\frac{\varepsilon}{2}.$$

Thus Theorem 2 and Lemma 7.1 give

(8.1) $$\mathcal{B}_{d}\leq Y^{7/12+(342/7)\varepsilon_{0}}.$$

By (2.3) at scale $Y$, at a further cost $Y^{\varepsilon_{0}}$, $N^{*}_{\mu}(Y)\ll_{\varepsilon}Y^{7/12+\varepsilon/2}$. The constant depends only on $\varepsilon$. Finally (2.1) with $\lambda=1$ absorbs $(\log X)^{2}$ and the maxima into the remaining $\varepsilon/2$. ∎

Appendix A The thirteen min-disjunctions

We work at scale $X$ with exponent $\lambda$; Lemma 7.2 uses the same derivation at scale $Y$ with $\mu$. Put $p_{i}=a_{i}+b_{i}+c_{i}$, $L=a+b+c$. Each row is [2, Proposition 4.1] for one triple $I,I^{\prime},I^{\prime\prime}\subseteq[6]$. A sum of two terms is at most twice the larger, so

$$D\leq\max(\lambda+4\varepsilon_{0}-P,\lambda+6\varepsilon_{0}-Q)+O_{\varepsilon_{0},d}(1/\log X),$$

where $P=\sum_{i\in I}a_{i}+\sum_{i\in I^{\prime}}b_{i}+\sum_{i\in I^{\prime\prime}}c_{i}$ and $Q=1-\sum_{i\in I}(i-1)a_{i}-\sum_{i\in I^{\prime}}(i-1)b_{i}-\sum_{i\in I^{\prime\prime}}(i-1)c_{i}$. With $C_{\varepsilon_{0},d}$ the constant of the source estimate, the constant and the factor $2$ cost $\log(2C_{\varepsilon_{0},d})/\log X=O_{\varepsilon_{0},d}(1/\log X)$. By the lower bound in (2.4) and (2.5),

$$2^{-O_{d}(1)}X^{\lambda}\leq\prod_{j\leq d}A_{j}B_{j}C_{j}=X^{L},\qquad\lambda\leq L+O_{\varepsilon_{0},d}(1/\log X).$$

Hence $\min(P,Q)\leq L-D+6\varepsilon_{0}+O_{\varepsilon_{0},d}(1/\log X)$. The certificate uses the limiting form below, free of $\varepsilon_{0}$; the slack $6\varepsilon_{0}+O_{\varepsilon_{0},d}(1/\log X)$ is charged in Lemma 7.2 and Table 7.1. The choices, from the proof of [2, Theorem 1.3]:

$$\begin{array}[]{ll}\text{row }1:&I=I^{\prime}=I^{\prime\prime}=\{1,2,4\};\\ \text{rows }2\text{--}4:&I=I^{\prime}=\{1,2\},\ I^{\prime\prime}=\{1,2,3\}\ \text{and permutations};\\ \text{rows }5\text{--}7:&I=\{1,2\},\ I^{\prime}=I^{\prime\prime}=\{1,2,3\}\ \text{and permutations};\\ \text{rows }8\text{--}10:&I=I^{\prime}=\{1,2\},\ I^{\prime\prime}=\{1,3\}\ \text{and permutations};\\ \text{rows }11\text{--}13:&I=\{1,2\},\ I^{\prime}=I^{\prime\prime}=\{1,3\}\ \text{and permutations}.\end{array}$$

Written out:

(G1) $$P_{1} =p_{1}+p_{2}+p_{4}, Q_{1} =1-p_{2}-3p_{4}$$

(G2) $$P_{2} =p_{1}+p_{2}+a_{3}, Q_{2} =1-p_{2}-2a_{3}$$

(G3) $$P_{3} =p_{1}+p_{2}+b_{3}, Q_{3} =1-p_{2}-2b_{3}$$

(G4) $$P_{4} =p_{1}+p_{2}+c_{3}, Q_{4} =1-p_{2}-2c_{3}$$

(G5) $$P_{5} =p_{1}+p_{2}+a_{3}+b_{3}, Q_{5} =1-p_{2}-2a_{3}-2b_{3}$$

(G6) $$P_{6} =p_{1}+p_{2}+b_{3}+c_{3}, Q_{6} =1-p_{2}-2b_{3}-2c_{3}$$

(G7) $$P_{7} =p_{1}+p_{2}+c_{3}+a_{3}, Q_{7} =1-p_{2}-2c_{3}-2a_{3}$$

(G8) $$P_{8} =p_{1}+a_{3}+b_{2}+c_{2}, Q_{8} =1-2a_{3}-b_{2}-c_{2}$$

(G9) $$P_{9} =p_{1}+b_{3}+a_{2}+c_{2}, Q_{9} =1-2b_{3}-a_{2}-c_{2}$$

(G10) $$P_{10} =p_{1}+c_{3}+a_{2}+b_{2}, Q_{10} =1-2c_{3}-a_{2}-b_{2}$$

(G11) $$P_{11} =p_{1}+a_{2}+b_{3}+c_{3}, Q_{11} =1-a_{2}-2b_{3}-2c_{3}$$

(G12) $$P_{12} =p_{1}+b_{2}+a_{3}+c_{3}, Q_{12} =1-b_{2}-2a_{3}-2c_{3}$$

(G13) $$P_{13} =p_{1}+c_{2}+a_{3}+b_{3}, Q_{13} =1-c_{2}-2a_{3}-2b_{3}$$

and each row is used as the disjunction $D\leq L-P_{j}$ or $D\leq L-Q_{j}$.

Appendix B Computer checks

External inputs, supplied with the paper: the certificate tree, its leaf scan, the stability scan, the two witness scans. Everything else is in the text, including the witness coordinates of Section 6.1.

The displayed system alone gives the $8192$ cells, the cut-free optimum $3/5$, and the root-cell scan: $7768$ cells need no cut, $424$ do. The file cert_leaves_tail.jsonl gives the rest. It holds the cut descriptors and their order on paths, $10\,820$ leaves, $2628$ internal nodes, $732$ distinct cuts among the $1\,560\,195$, and the multipliers. All arithmetic is exact. No program imports a generator, a numerical package or an LP solver.

Format: one metadata record, then one record per leaf with exactly the fields cell, path, kind, lam.

• cell: a list in $\{1,2\}^{13}$; entry $j$ (zero-based) selects ("dis",j,cell[j]).

• path: a list of [cut-descriptor,sign] pairs.

• lam: a sparse list of [row-key,"p/q"], multipliers positive.

• kind: always "bound"; only the fourth program reads it.

A cut descriptor is [sigma,R_y,R_z,T]: $R_{y},R_{z}$ increasing subsets of $[6]$, not both empty; $T$ a nonempty increasing subset of $\{1,\ldots,6,\mathrm{tail}\}$; $(y,z)=(b,c),(a,c),(a,b)$ for $\sigma=a,b,c$. The quantities $q,C_{T},W_{T},F,\mathsf{A},\mathsf{B},\tau_{s}$ are those of (5.1), (5.2) and after (5.3). A path step is a descriptor and a sign, e.g. [["b",[3],[3],[4,7]],"+"].

Admissible rows at a leaf in cell $\mathcal{C}$: the $43$ linear rows; the thirteen rows selected by $\mathcal{C}$; the rows its path authorizes, namely cutP for a $+$ step and cutMA, cutMB for a $-$ step. Nothing else is admissible. The text defines every row and (6.2). It fixes neither the tree nor a numbering of cuts; the file names each cut by its descriptor. The four programs below are the certificate checks; the statistics scripts are not.

program	what it checks
verify_712_independent.py	Tables 6.1 and B.2; each leaf’s rows are valid on its cell and path; (6.2) at every leaf; that the declared target is $7/12$ and that no leaf’s certificate value exceeds it; the points $x(0)$ and (6.5) of Section 6.1 against all $1\,560\,195$ instances of (5.3). It also runs four smoke tests, corruptions it must reject: a malformed cell, a missing branch (a deleted sibling leaf), a negative multiplier and a sign-mismatched row. Soundness rests on (6.2) at every leaf with the admissibility and complete-trie checks of Section 6.
verify_cert_tail_independent.py	the same cell and leaf structure and the same Farkas conditions, from a second implementation sharing no code with the first
check_stability_constant.py	$\Lambda=18/7$ of Lemma 7.1, and the largest number of nonzero multipliers on a leaf; it reads the multipliers only and is not a validity check
blind_fourth/verify_blind_third.py	a code-independent reconstruction from the displayed formulas, checking row authorization, complete binary path tries, the Farkas conditions and the stability scan; its self-test rejects malformed certificates

Table B.1 is generated from the verifier’s row labels. Here $s\in\{a,b,c\}$, $(x,y)\in\{(a,b),(a,c),(b,c)\}$, $(i_{a},i_{b},i_{c})\in\{0,1,2\}^{3}$. The branch tag is $b\in\{1,2\}$, not the side $b$. The program indexes the tail by $7$.

Table B.1. Verifier row labels and their paper sources.

verifier label	definition and source
("L",)	(2.8), from [2, (6.1)]
("62",s)	(2.9), from [2, (6.2)]
("63",s)	(2.10), from [2, (6.3)]
("T7",s)	$v_{s}\geq 7\tau_{s}$, the second row of (3.1)
("TV",s)	$\sum_{i\leq 6}is_{i}+v_{s}\leq 1$, the third row of (3.1)
("64",x,y)	(2.11) for the pair $(x,y)$, from [2, (6.4)]
("F",i_a,i_b,i_c)	(2.12), where on side $s$ the index $i_{s}=0,1,2$ selects respectively $s_{2}+s_{4}+s_{6}$, $s_{3}+s_{6}$, and $s_{5}$, from [2, (6.10)]
("dis",j,b)	$0\leq j\leq 12$: the $P$ arm of $(G_{j+1})$ if $b=1$, and the $Q$ arm if $b=2$. The order is $(6.5)$, $(6.6)$a/b/c, $(6.7)$ab/bc/ca, $(6.8)$a/b/c, $(6.9)$a/b/c.
("cutP",sigma,R_y,R_z,T)	$D\leq\mathsf{A}+\mathsf{B}$, the plus child of (6.1).
("cutMB",sigma,R_y,R_z,T)	$\mathsf{B}\leq 0$, the sign row of the minus child.
("cutMA",sigma,R_y,R_z,T)	$D\leq\mathsf{A}$, the bound row of the minus child.

In the disjunction row, the exact index correspondence is ("dis",j,b) $\leftrightarrow(G_{j+1})$ for $j=0,\ldots,12$. The verifier stores $e_{2}=1-Q_{j+1}$, since its second branch is $1-e_{2}\leq L-D$.

All $2^{13}=8192$ arm combinations are represented, with no pruning among the thirteen min-disjunctions. Every cell is non-empty, so none is infeasible; some cells contain others and are checked all the same. Within each cell (6.1) is applied to selected cut rows.

In each cell the paths must form a complete binary trie, as in the proof of Theorem 2. Duplicate paths and prefix leaves are rejected. Hence $10\,820=8192+2628$, as for any full binary forest with $8192$ trees. Table B.2 gives further statistics; the proof uses none of them.

Table B.2. Further statistics of the certificate.

largest number of distinct cuts in one cell	$35$
largest number of leaves in one cell	$42$
largest path depth	$13$
distinct multiplier vectors	$4545$
largest number of nonzero multipliers on one leaf	$21$
distinct multiplier denominators	$4009$
largest multiplier denominator	$29\,322\,517$
nonzero multipliers in the whole certificate	$126\,499$
each of the thirteen disjunctions used in both branches	yes
second-worst certificate value	$9154/15693$

Table B.3. SHA-256 of the files this paper depends on.

file	SHA-256
cert_leaves_tail.jsonl	0a92f82f24e3dc4bca87f4592633ba
	ceebaa6a91423048561deb23b7031fd5e7
verify_712_independent.py	2e40f9527368c3d85b5553615f2445f
	81409bfe4c0a68267646d4e41ac4a2b75
verify_cert_tail_independent.py	226e2ed54b620983df5dd311fae529
	3d40e33055ce1270cb2637c8910d5059fe
check_stability_constant.py	0e491fb351fafebb4445268a29703bd9
	ec00fced4a56db64a4c4d4d04c582451
blind_fourth/verify_blind_third.py	e733c750ce60e64236008d185e1725e0
	83cb7961f58b7868117428c8a989f316

Statements and declarations

Competing interests. The author has no competing interests.

Funding. No funding was received for this work.

Data availability. The certificate, the four checking programs of Appendix B and their outputs accompany this paper; their SHA-256 values are in Table B.3.