Ball and Simoens spent the proof getting down to twelve graphs; the computer needed a minute on each

Three rules on the pattern

A quantum Latin square of order six is a six-by-six array whose every row and every column gives an orthonormal basis of six-dimensional complex space, and two such squares count as orthogonal when the 36 tensor products of their matching entries form an orthonormal basis of the product space. Ball and Simoens move the argument onto the pattern of those vectors: write a one in place of every nonzero coordinate and each entry becomes a binary string. Three rules then apply. Standard form fixes the first row of both squares. Because every row and column of a quantum Latin square gives a unitary pattern, an entry of weight one imposes zeros on every entry in its row and column. The third rule says that from the second row onward, each one in a square forces a zero at the same position in the orthogonal square. The chain of lemmas built on those three rules ends at Theorem 20: if two mutually orthogonal quantum Latin squares of order six exist at all, one of them can be taken to be an ordinary classical Latin square.^[1]

The weight of the proof sits on that theorem. Before it, the question wanders through a continuous space: infinitely many choices of unit vector in six complex dimensions. After it, one of the two squares becomes a finite object drawn from a finite list, and what remains is a question about the other square alone. The rules that get there are plain enough to state in a line and each does real work; it is the third rule that converts orthogonality between the two squares into a constraint on individual positions. Another reading is available: the reduction may be a convenience rather than a necessity, and a different formulation might have carried the continuous search directly. The paper reports no attempt without the reduction, so that comparison is not on the table.^[1]

Ten by machine, two by hand

Lemma 21 turns the remaining question into graph theory. Take the Latin square graph of the classical square: its vertices are the 36 entry positions, and two vertices are joined when they share a row, a column or a symbol. Two mutually orthogonal quantum Latin squares of order six then exist exactly when the complement of such a graph admits an orthonormal representation in six complex dimensions. Latin squares that turn into one another under permutation of the roles of rows, columns and symbols share a graph, and under that equivalence there are only twelve Latin squares of order six; the classification has been known since Schonhardt. The whole question therefore comes down to twelve graphs. The algorithm the pair wrote runs for about a minute on each.^[1]

The algorithm answers in one direction only. When it returns False, no orthonormal representation exists and that case is closed; when it returns True, nothing follows. Ten of the twelve graphs return False. Both of the remaining two contain a subsquare of order three, and they are finished with a hand argument ending at Theorem 26. Ball and Simoens remark in passing that one further dependency case in the code would have resolved one more graph, and that they left it out to keep the code simple. The split into ten and two therefore marks where the authors stopped extending the search; the mathematics draws no line there.^[1]

What this closes

What settles is narrow and exact. Rather and colleagues built an entangled solution to the thirty-six officers problem in 2022, and the question left open afterwards was whether the same square could be filled by superposition alone, in the sense of orthogonality that Musto defines. It cannot. Życzkowski, from the team behind the 2022 construction, reads the new result as showing their own solution cannot be made simpler. Jamie Vicary's observation that absolutely maximally entangled states can serve as error-correcting codes in quantum computers points at a possible use; this proof establishes nothing about that use. The refereed version appeared in Physical Review Letters, while the text this column works from is the open preprint posted to arXiv.^[1]

Simoens says that he and Ball are now looking at the seven-by-seven case. The reduction will not carry over there unchanged: its finiteness comes from a classification specific to order six, the twelve main classes. If a result at order seven arrives along the same route, the paper will have to report how many main classes it enumerates and how many of them the algorithm closes. That is the number to watch, and it can be expected before the end of 2028.^[1]