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Take a 3 × 3 grid and fill it with 1, 2 and 3 so that no number repeats in any row or column. That is a Latin square, the idea behind Sudoku. Now ask a harder question: can two such squares be laid on top of each other so that every pair of symbols appears exactly once? For some sizes the answer is yes. For others, Leonhard Euler guessed it was no, and for 177 years nobody could settle the question in general. An Indian statistician, Raj Chandra Bose, was part of the team that settled it. This post explains the problem with a worked example and tells his story.
What is an orthogonal pair of Latin squares?
The 1960 paper by Bose, Shrikhande and Parker in the Canadian Journal of Mathematics gives the definitions. A Latin square of order v is an arrangement of v symbols in a v × v square so that each symbol occurs exactly once in every row and once in every column. Two Latin squares are orthogonal if, when they are superposed, each symbol of the first occurs just once with each symbol of the second.
Worked example, order 3. Square A uses 1, 2, 3; square B uses a, b, c.
A, row by row: (1, 2, 3), (2, 3, 1), (3, 1, 2)
B, row by row: (a, b, c), (c, a, b), (b, c, a)
Lay B on A and read the pairs.
Row 1: 1a, 2b, 3c
Row 2: 2c, 3a, 1b
Row 3: 3b, 1c, 2a
There are 3 × 3 = 9 possible pairs, and all 9 appear exactly once. So A and B are orthogonal. Check each square is Latin too: every row and column of A has 1, 2, 3 once, and the same holds for B with a, b, c.
What did Euler conjecture?
Euler's 1782 memoir on a new kind of magic square is reference 8 in the 1960 paper, which states his conjecture: no two orthogonal Latin squares exist of order v = 4t + 2. Those orders are 2, 6, 10, 14, 18, 22 and so on.
The order 6 case is the famous 36 officers problem. According to MacTutor's biography of Gaston Tarry, six regiments each send six officers of six different ranks, and the question is whether they can stand in a 6 × 6 square with no rank or regiment repeated in any row or column. Tarry proved that two orthogonal Latin squares of order 6 do not exist. So Euler was right about 6. Order 2 is easy to check by hand: there are only two Latin squares of order 2 and they cannot be orthogonal.
How was the conjecture disproved?
The 1960 paper tells the sequence. Parker first showed that a related conjecture by MacNeish was false, which threw serious doubt on Euler's. Then Bose and Shrikhande, in a 1959 PNAS paper titled "On the falsity of Euler's conjecture", obtained a counterexample using pairwise balanced designs. They went on to show the conjecture fails for infinitely many orders. Parker then used a method of differences to construct an orthogonal pair of order 10, in his own 1959 PNAS paper.
Finally, the three together proved that Euler's conjecture is false for all v = 4t + 2 greater than 6. Every order in Euler's list fails except 2 and 6.
MacTutor's biography of R. C. Bose adds that the result reached the front page of the Sunday New York Times on 26 April 1959, and that the three were nicknamed "Euler's spoilers".
Who was R. C. Bose?
MacTutor records that Bose was born on 19 June 1901 in Hoshangabad and grew up in Rohtak. His mother died in the 1918 influenza epidemic, and money was often short: after his MA in pure mathematics in Calcutta in 1927 he worked as a lecturer so poorly paid that he took on private tuition. In 1932 P. C. Mahalanobis offered him a part-time post at the Indian Statistical Institute. He did not take a doctorate until 1947, after he already had many published papers. He moved to the University of North Carolina at Chapel Hill in 1949 and later to Colorado State University, where he died on 31 October 1987.
His name lives in coding theory too. With D. K. Ray-Chaudhuri he published On a class of error correcting binary group codes in 1960; MacTutor notes these are the BCH codes, named partly after him.
Why does this matter outside puzzles?
Bose came to Latin squares through statistics. The 1960 paper links orthogonal Latin squares to block designs and orthogonal arrays, tools for planning experiments so that every combination of conditions is tested fairly. If a farm trial compares fertilisers across fields and seasons, an orthogonal design makes sure no fertiliser is always paired with the same field.
What can a student take from Bose's way of working?
Test the small cases yourself. Building an order-3 pair by hand is how you understand the definition.
Distrust a pattern that has only been checked a few times. Euler was right for 2 and 6 and wrong for every larger case.
Bring tools across subjects. Designs from agricultural statistics broke a problem in pure mathematics.
If counting arguments trip you up, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see why each count works. Our guide to permutations and combinations covers the counting behind these grids, and our post on Leonhard Euler shows the rest of his work.
Your next step: build two orthogonal Latin squares of order 4 using symbols 1-4 and a-d, and list all 16 pairs to check.
