The difference-matrix property gives a unique \(c\), then (1) gives a unique \(x\). Hence \(L_r,L_s\) are orthogonal. Therefore an \((n,k,1)\)-DM gives \(k-1\) MOLS of order \(n\).

4. Independent reconstruction of nine MOLS(108)

The 2025 construction uses the additive group

Only addition is needed, so the checker represents these as the vector spaces

\(\mathbb F_2^2\) and \(\mathbb F_3^3\); no finite-field package or irreducible-polynomial implementation is trusted.

The paper supplies ten-row, twelve-column seed data \(F=[F_1\mid F_2\mid F_3]\) and ten-entry vectors \(V,W\). Its 108 columns are

The literal seed entries are in the companion script, transcribed from arXiv:2412.00480.

(d) The dependency-free base run checks:

• all \(\binom{10}{2}=45\) row-pair difference multisets, each against all 108 group elements;
• all 972 rows and all 972 columns of the nine resulting \(108\times108\) squares;
• all \(\binom92=36\) square pairs, checking 11,664 distinct ordered symbol pairs in each.

It reports these deterministic certificate hashes (using the element and row order fixed in the script):

SHA256 difference_matrix = 53f0e738f6bce9becb2dc538ca7eaa531340919a15e22011753c87a1292ae465
SHA256 nine_mols = 45c1035c09a7dde1a7d24323ea98d3828b0b26c4cd2acf38c9f6b35663e3b9ac

Combining this finite check with the elementary implication in Section 3 verifies \(f(108)\geq9\). It does not say \(f(108)=9\).

5. Exact obstruction in the natural equivariant extension family

The compact form (2) suggests trying an eleventh DM row with the same \(3^2\)-development:

where \(v,w\in\mathbb F_3^3\) and all twelve \(f_c\in G\) are free. If successful, this would give ten MOLS(108).

Fix an old row \(i\), and put

For a fixed seed column \(c\), the nine differences between (3) and row \(i\) form

(a) If \(\dim H_i<2\), (4) has at most three distinct values among nine developed columns, so a difference permutation is impossible. If \(\dim H_i=2\), (4) contains all nine elements of one coset exactly once. The complete set of 108 differences is a permutation of \(G\) if and only if the twelve seed columns select all twelve cosets of \(\{0\}\times H_i\).

For rank two, a nonzero normal

labels those cosets by

Thus the proposed row exists if and only if, simultaneously for all ten old rows \(i\),

A global translation of every \(f_c\) preserves all differences, so \(f_0=0\) loses no solutions.

This is an exact reduction from a nominal \(27^2\cdot108^{11}\) search to 729 small, finite CSPs. The program also directly enumerates every \(H_i\), checks the rank/cross-product equivalence, and verifies that every quotient-label fibre is exactly the corresponding nine-element coset; the two solvers therefore do not merely agree on an unchecked algebraic reduction.

(d) Exhaustive result:

| Quantity | Exact value |

|---|---:|

| \((v,w)\) pairs | 729 |

| Rejected because some \(\dim H_i<2\) | 618 |

| Rank-valid pairs | 111 |

| Rank-valid CSPs satisfiable | 0 |

| Dependency-free backtracking nodes, total | 102,728 |

| Backtracking wall time | 46.920 s |

The backtracker uses minimum-remaining-values branching, forward checking, and ten 12-bit used-label masks. It considers every possible value in \(G\) for each unfixed \(f_c\). As a second enumeration engine, OR-Tools CP-SAT 9.15.6755 modeled the same 111 all-different CSPs and returned INFEASIBLE for every one, with no timeout or UNKNOWN; the combined audit took 138.591 seconds on one worker.

The exact conclusion is:

> (d) The published \((108,10,1)\)-difference matrix has no eleventh row of form (3).

It does not rule out:

• another, non-equivariant row extending this DM;
• a different \((108,11,1)\)-difference matrix;
• ten MOLS(108) not arising from a difference matrix;
• or the asymptotic square-root conjecture.

6. Unrestricted extension probe and measured computation wall

An arbitrary common orthogonal mate \(X\) of the verified nine squares can be modeled with one variable \(X_{r,c}\in\{0,\ldots,107\}\) per cell. Impose all-different constraints on:

1. every row of \(X\);

2. every column of \(X\);

3. for every old square \(L_i\) and symbol \(s\), the 108 cells on which \(L_i=s\).

(a) These constraints are exactly the Latin and nine orthogonality conditions. Relabeling the symbols of \(X\) permits its first row to be fixed without losing a solution.

(d) The resulting CP-SAT model has 11,664 integer variables, 1,188 all-different constraints, and 108 fixed first-row entries. A one-worker run with a nominal 90-second limit ended after 105.942 wall seconds because presolve overran the limit. Presolve expanded it to:

1,132,488 Boolean variables
138,672 exactly-one constraints
13,589,856 literals

The status was UNKNOWN, with zero conflicts and zero branches: it never reached search. This conveys no mathematical evidence for or against an extension.

(c) A longer run would need to solve or certify this million-Boolean instance, but the pilot gives no branch rate from which a defensible completion time could be extrapolated. The closest published generic benchmark is already about \(10^{20}\) years for the much smaller open 3-MOLS(10) search described in Section 2.3. Consequently, an unrestricted order-108 computation is not responsibly budgetable as a few-core-hour job. The exact model is included as an optional, off-by-default probe so that a future solver or symmetry reduction can be tested reproducibly.

7. Exact bottleneck in the standard asymptotic machinery

Suppose Wilson's recurrence (W) is asked to prove

for some fixed \(c>0\). Since the elementary upper bound is \(f(s)\leq s-1\), every term in the minimum forces

in any non-circular use of (W). (The convention \(f(0)=f(1)=\infty\) permits \(u\in\{0,1\}\); if \(u\geq2\), it additionally forces \(u\geq K+1\). Taking \(t=1\) merely puts \(f(n)\) itself back into the minimum and gives no new bound.)

Together with \(n=mt+u\) and \(u\leq t\), (6) forces

and if \(u\geq2\), then \(u=\Theta(\sqrt n)\) as well. Moreover (W) requires \(f(m)\) and \(f(m+1)\) both to be at least \(K\). Since \(m,m+1=\Theta(\sqrt n)\), this is a linear-in-the-suborder lower bound at two consecutive orders, as well as at \(t\) and at any nontrivial \(u\).

(a) This is the precise square-root bottleneck of (W): it cannot bootstrap merely sublinear information at smaller orders to the desired bound. It needs linear-size MOLS at three smaller orders, including a consecutive pair, and normally at a fourth order \(u\).

The MacNeish bound used inside the classical sieve route is

If (M) alone is to certify these bounds, every prime-power component of each of \(m,m+1,t\), and of \(u\) when \(u\geq2\), must be at least \(K+O(1)=\Theta(\sqrt n)\). But each such whole number is only \(O(\sqrt n)\). For sufficiently large \(n\), it therefore has only one prime-power component: each is itself a prime power.

(a) Hence the exact missing number-theoretic input for the unmodified Wilson--MacNeish route would be a uniform representation

with \(m,m+1,t=\Theta(\sqrt n)\) all prime powers and either \(u\in\{0,1\}\) or \(u=\Theta(\sqrt n)\) another prime power (or comparably strong non-MacNeish lower bounds at those orders). No such lemma is supplied by the cited sieve results. A genuinely promising advance must either provide this simultaneous linear information or replace the recurrence so that the consecutive-order minimum disappears.

(b) Modern enumeration theorems do not fill this gap: the lower enumeration asymptotics used by Boyadzhiyska--Das--Szabó require fixed \(k\), while the desired \(k\asymp\sqrt n\) grows with \(n\). Their paper explicitly notes that existing lower-counting methods do not presently extend cleanly when \(k\) grows.

This identifies the wall more precisely than “the problem is open”: the known sieve/recursion architecture loses its power exactly because the square-root target turns every nontrivial recursive subproblem into a linear-density MOLS problem.

8. Reproduction

The standalone checker is erdos724_wave7i_verify.py. It uses only the Python 3.12 standard library for the certificate and exhaustive backtracker. OR-Tools is optional and is needed only for the second solver and unrestricted probe.

Base certificate check (about 0.24 seconds here):

python runs/erdos724_wave7i_verify.py

Dependency-free exhaustive structured search (46.920 seconds here):

python runs/erdos724_wave7i_verify.py \
  --search-extension \
  --extension-solver backtrack

Independent CP-SAT audit as well (about 139 seconds here):

python runs/erdos724_wave7i_verify.py \
  --search-extension \
  --extension-solver both \
  --per-candidate-seconds 10

Bounded unrestricted probe (expect UNKNOWN; detailed presolve log enabled):

python runs/erdos724_wave7i_verify.py \
  --probe-arbitrary-extension \
  --arbitrary-time-limit 90 \
  --cp-workers 1 \
  --cp-log

The final script's SHA-256 is:

11d55db1ed4e94f43e6f74c56f916a552d11e6b59917398a0eb93e1bf9d53fc1

PARTIAL: corroborated Lu's stronger \(f(n)\geq n^{10/143}-2\) literature bound, independently verified nine MOLS(108), and exhaustively ruled out an eleventh row in the published matrix's natural equivariant family; the uniform \(\Omega(\sqrt n)\) question remains open.