Selected Topics and Applications
The Refutation of Euler's Conjecture
The 1959–60 disproof of Euler's conjecture by Bose, Shrikhande and Parker, and the algebraic construction that produced the counterexamples.
Learning objectives
- Describe the counterexample of order 10 and the general result
- Follow the algebraic strategy behind the construction
- Assess what the episode shows about applied universal algebra
The result
A pair of orthogonal Latin squares of order n exists for every n except n = 2 and n = 6.
Euler's conjecture is therefore false for every order congruent to 2 mod 4 beyond 6. The first counterexample, of order 10, was found by Parker; Bose and Shrikhande produced order 22 and then, jointly with Parker, the general result.
Orders 2 and 6 are the only failures, and both are verifiable by finite search — order 2 trivially, order 6 by Tarry's 1901 enumeration. Every other order admits an orthogonal pair. The conjecture was wrong everywhere it made a non-trivial claim.
The strategy
The product construction alone cannot reach orders congruent to 2 mod 4, because such an order has exactly one factor of 2 and order 2 admits no orthogonal pair. The essential new ingredient is the patching step, which allows a block of the constructed square to be replaced independently.
The formal setting for the patching argument is the theory of pairwise balanced designs, developed by Bose and Shrikhande for exactly this purpose. It provides a systematic bookkeeping for which orders can be assembled from which ingredients.
What the algebraic view contributed
Products
The coordinatewise construction is immediate once orthogonality is stated as a condition on operations. Combinatorially it requires explicit verification.
Subsquares as subalgebras
The patching step replaces a subquasigroup by another of the same order. Framing subsquares as subalgebras makes the legitimacy of the replacement clear.
Systematic search
Reformulating the problem algebraically turned an apparently unstructured search into a construction problem with known building blocks.
The source presents this episode as evidence for its claim that applied universal algebra would grow in importance. The claim was that algebraic framing is not decoration but the thing that makes the construction visible.
Assessment
It would overstate the case to say universal algebra solved the problem. The decisive ingredients were combinatorial designs and substantial explicit construction. What the algebraic framing supplied was the product and substitution operations, and a language in which the pieces compose.
The honest summary is that the algebraic viewpoint made the composition structure of the problem visible, and that composition structure is what the eventual proof exploits. That is a real contribution without being the whole of the proof.
When a combinatorial class is closed under products and admits substructures, recasting it as a variety supplies constructions for free. Steiner systems, Latin squares, and finite automata in the next pages all follow this pattern.
Frequently asked questions
Why did the conjecture survive 177 years?
Because the two smallest cases both failed, no construction was known for any order congruent to 2 mod 4, and Tarry's exhaustive verification of order 6 gave the pattern strong empirical support. The counterexamples require constructions considerably more elaborate than anything available in Euler's time.
Are there three mutually orthogonal Latin squares of order 10?
Yes. Whether there are nine — the maximum, equivalent to a projective plane of order 10 — was settled negatively by a large computer search in 1989.
Source. S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, The Millennium Edition — a corrected re-typesetting of Springer-Verlag Graduate Texts in Mathematics 78 (1981). Section III.3, book pages 115-118.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates and reorganises mathematical results; it is not a reproduction of the source text.