Abstract
A Latin square of order $n$ places $n$ symbols in an $n\times n$ grid so that each occurs once in every row and every column. Two Latin squares are orthogonal if, when superimposed, all $n^2$ pairs of symbols occur exactly once. Euler asked whether officers of six ranks from six regiments can stand so that every row and every column holds one officer of each regiment and each rank, which is to ask whether there is an orthogonal pair of order $6$. No new theorem or law is claimed.
First, with first row and column in order, there are $9408$ squares of order $6$——constructing all such squares gives $1,1,1,4,56,9408$ for orders $1$ to $6$. The full numbers without this ordering are $576$ for order $4$ and $161280$ for order $5$, both agreeing with counts made by constructing them all (Section 2).
Second, and this is the core. None of the $9408$ squares of order $6$ has an orthogonal mate——having a mate is the same as splitting the square into $n$ disjoint transversals (sets of $n$ cells taking one from every row, every column and every symbol). All $9408$ were checked, and $0$ could be split (Section 3).
Third, smaller orders have mates——the squares with a mate number $0$ in order $2$, $1$ of $1$ in order $3$, $1$ of $4$ in order $4$ and $6$ of $56$ in order $5$. In orders $3$ and $4$ the answer by transversals and the answer by trying every candidate mate agreed square by square (Section 3).
Fourth, the lack of a mate is not a lack of transversals——of the order-$6$ squares only $2100$ ($\mathbf{0.223214}$) have no transversal; the rest have $8$ ($7020$ squares), $24$ ($108$) or $32$ ($180$). Still, none can be split into $6$ disjoint ones (Section 3).
Fifth, the separator is not “even” but “six”——the square with $i+j$ modulo $n$ in row $i$ and column $j$ has $0$ transversals in even orders ($2$, $4$, $6$) and $3$, $15$, $133$ in odd orders. But in order $4$ another square, the table of the Klein four-group, has $8$ transversals and a mate. It is not that even squares all lack mates; only order $6$ lacks a mate for every square (Section 4).
Keywords: Latin square, orthogonal Latin squares, thirty-six officers problem, transversal, combinatorics.
Scope (what is not claimed)
No new theorem or law is claimed——Euler’s question ($1782$), that there is no orthogonal pair of order $6$ (Tarry, $1900$), that Euler’s conjecture fails from order $10$ on (Bose, Shrikhande and Parker, $1960$), and the numbers of small Latin squares (McKay, Meynert and Myrvold, $2007$) are all known. Tarry’s proof is not reproduced——the $9408$ squares with first row and column in order are constructed and checked one by one by transversals. Orders $10$ and above are not computed——the side on which Euler’s conjecture (no pair for orders leaving remainder $2$ on division by $4$) fails is stated only as a known result. Uses in the design of experiments are not treated. Relation to earlier papers: Paper 247 counted eight ways of “splitting into even and odd”——that cyclic squares of even order have no transversal is one split by parity, but the absence in order $6$ cannot be explained by it. What is added is giving the distribution of transversal counts over the $9408$ ordered squares of order $6$, matching the transversal method with exhaustive trial square by square in orders $3$ and $4$, and placing the separator on “six” rather than “even” and checking it by setting the cyclic squares of even order beside the Klein four-group table of order $4$.
1. Notation and the Question
Setting: a Latin square $L$ of order $n$ places the symbols $0,1,\dots,n-1$ in an $n\times n$ grid so that each occurs once in every row and every column. Two squares $L,M$ are orthogonal if all $n^2$ pairs $(L_{ij},M_{ij})$ occur exactly once. A transversal of $L$ is a set of $n$ cells, one from every row and every column, whose symbols are all different.
Mates and transversals: if $M$ is orthogonal to $L$, the $n$ cells where $M$ has the same symbol form a transversal of $L$, and $n$ of them split all the cells. Conversely, if $L$ can be split into $n$ disjoint transversals, placing one symbol per transversal gives $M$.
Why ordering is allowed: permuting the rows of $L$ and $M$ in the same way, or their columns in the same way, keeps them orthogonal. Every square can be brought, by permuting columns, to have first row $0,1,\dots,n-1$, and then, by permuting the other rows, to have first column $0,1,\dots,n-1$ too. So whether a mate exists can be decided on squares with first row and column in order (reduced squares) alone.
Question: does a square of order $6$ have an orthogonal mate? If not, is that because it has no transversal, or because $6$ is even?
2. Constructing All the Squares
Computation 2.1: all reduced squares were constructed by a search that fills one cell at a time.
| order | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| reduced squares | $1$ | $1$ | $1$ | $4$ | $56$ | $9408$ |
Computation 2.2: constructing and counting all squares without the ordering gave $576$ for order $4$ and $161280$ for order $5$. These agree with $n!(n-1)!$ times the reduced number: $4! 3!\times4=576$ and $5! 4!\times56=161280$. By the same product, order $6$ has $6! 5!\times9408=812851200$ squares in all.
Observation 2.3 the two counts agree. The reduced squares and the full squares were each constructed separately, and the product relation held in orders $4$ and $5$——$9408$ is also the value given by McKay and coauthors.
3. Looking for Mates
Computation 3.1: for every reduced square, all transversals were counted, and whether it splits into $n$ disjoint transversals (has a mate) was checked.
| order | reduced squares | with a mate | transversals (number of squares with that many) |
|---|---|---|---|
| $2$ | $1$ | $0$ | $0$ ($1$) |
| $3$ | $1$ | $1$ | $3$ ($1$) |
| $4$ | $4$ | $1$ | $0$ ($3$), $8$ ($1$) |
| $5$ | $56$ | $6$ | $3$ ($50$), $15$ ($6$) |
| $6$ | $9408$ | $0$ | $0$ ($2100$), $8$ ($7020$), $24$ ($108$), $32$ ($180$) |
The $6$ squares of order $5$ with a mate coincided exactly with the $6$ having $15$ transversals.
Computation 3.2: independently of the transversal method, in orders $3$ and $4$ every full square ($12$ of order $3$, $576$ of order $4$) was superimposed to see whether any is orthogonal. The reduced squares with a mate were $1$ in order $3$ and $1$ in order $4$, agreeing square by square with the transversal method.
Observation 3.3 this is the core: in order $6$ no square has a mate. The thirty-six officers cannot be arranged, however they are placed——this is what Tarry showed in $1900$; this paper confirmed it by checking the $9408$ reduced squares one by one.
Observation 3.4 the lack of a mate is not a lack of transversals. Only $2100$ squares of order $6$ (share $\mathbf{0.223214}$) have no transversal; $7308$ have $8$ or more, and the most is $32$——there are enough transversals, yet $6$ mutually disjoint ones cannot be picked out. The issue is not how many, but how they fit together.
4. The Separator
Computation 4.1: transversals were counted for the square with the remainder of $i+j$ on division by $n$ in row $i$ and column $j$ (the cyclic square). For orders $2$ to $7$ they were $0,3,0,15,0,133$.
Observation 4.2 cyclic squares of even order have no transversal. Orders $2$, $4$ and $6$ give $0$, while odd orders give $3$, $15$ and $133$——so far it looks like a phenomenon split by even and odd.
Observation 4.3 the separator is not “even” but “six”. The only reduced square of order $4$ with a mate is the table adding $i$ and $j$ in binary without carries (the table of the Klein four-group), which has $8$ transversals. A square other than the cyclic one has a mate even in an even order——order $4$ has a square with a mate and order $6$ has none. Apart from order $2$, only order $6$ lacks a mate for every square. Euler conjectured that this extends to every order leaving remainder $2$ on division by $4$, but it fails from order $10$ on (Bose, Shrikhande and Parker).
Observation 4.4 beside Paper 247. Paper 247 counted eight ways of “splitting into even and odd”——that cyclic squares of even order have no transversal may be counted as one such split. But the absence in order $6$ is not a split by parity; order $4$ is even and has a mate.
Observation 4.5 what is not said. Tarry’s proof is not reproduced, orders $10$ and above are not computed, and uses in the design of experiments are not treated——what is shown is that none of the $9408$ reduced squares of order $6$ has a mate, the distribution of transversal counts, and that “even” is not the separator.
5. Roll Call and Closing
- reduced squares $1,1,1,4,56,9408$ (Computation 2.1): constructed in full.
- $576$ for order $4$ and $161280$ for order $5$ match the product (Computation 2.2): two counts.
- order $6$ has $812851200$ squares in all (Computation 2.2): $9408$ reduced ones suffice.
- $0$ squares of order $6$ with a mate (Computation 3.1): the core.
- $1$, $1$ and $6$ squares with mates in orders $3$, $4$, $5$ (Computation 3.1): smaller orders have them.
- transversals and exhaustive trial agree square by square in orders $3$ and $4$ (Computation 3.2): two routes.
- $2100$ squares of order $6$ without a transversal ($\mathbf{0.223214}$) (Observation 3.4): the rest have $8$ or more.
- transversals exist but do not split into $6$ (Observation 3.4): how they fit, not how many.
- cyclic squares of even order have $0$ transversals (Computation 4.1): it looks like a parity split.
- the Klein four-group table of order $4$ has a mate (Observation 4.3): the separator is “six”.
Closing: Euler asked whether officers of six ranks from six regiments can stand so that every row and every column holds one of each regiment and each rank. No Latin square of order $6$ has an orthogonal mate——constructing all $9408$ squares of order $6$ with first row and column in order, and checking each for a split into $6$ disjoint transversals, none could be split. Orders $3$, $4$ and $5$ have squares with mates, and in orders $3$ and $4$ the transversal method and exhaustive trial agreed square by square. The lack of a mate is not a lack of transversals——only $2100$ squares of order $6$ have none; most have $8$ or more, up to $32$, yet $6$ disjoint ones cannot be picked out. The separator is not “even” but “six”——cyclic squares of even order have no transversal, but in order $4$ the Klein four-group table has a mate. Placed among the earlier papers——beside the parity splits counted in Paper 247 it sets the absence in order $6$, which parity cannot explain. To be honest——Tarry’s proof was not reproduced, only confirmed by exhaustive trial of the reduced squares, and that Euler’s conjecture fails from order $10$ on was not computed.
References
- Bose, R. C., Shrikhande, S. S., & Parker, E. T. (1960). Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler’s conjecture. Canadian Journal of Mathematics.
- Euler, L. (1782). Recherches sur une nouvelle espèce de quarrés magiques. Verhandelingen uitgegeven door het zeeuwsch Genootschap der Wetenschappen te Vlissingen.
- McKay, B. D., Meynert, A., & Myrvold, W. (2007). Small Latin squares, quasigroups, and loops. Journal of Combinatorial Designs.
- Tarry, G. (1900). Le problème des 36 officiers. Compte Rendu de l’Association Française pour l’Avancement des Sciences.
- Yamagishi, Y. (2026). Splitting by Parity Is Not One Phenomenon [Paper 247]. Zenodo. 10.5281/zenodo.22159832
Reference
Yamagishi, Y. (2026). No Latin Square of Order Six Has an Orthogonal Mate [Paper 604]. YOI: https://www.yamagishiyuuki.com/yoi/604
DOI: 10.5281/zenodo.23014438
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 604. This paper is a preprint and has not been peer reviewed.