XLVIII OM - II - Problem 4

Determine all triples of positive integers having the following property: the product of any two of them gives a remainder of $ 1 $ when divided by the third number.
Let $ a $, $ b $, $ c $ be the numbers we are looking for. From the given conditions, these numbers are greater than $ 1 $ and are pairwise coprime. The number $ bc + ca + ab - 1 $ is divisible by $ a $, by $ b $, and by $ c $, so it is also divisible by the product $ abc $:

After dividing both sides by $ abc $:

The sum on the left side of equation (1) is greater than $ 1 $, so $ k > 0 $. The sum on the left side of (2) is less than $ 2 $, so $ k < 2 $. Therefore, $ k = 1 $. The right side of (2) is greater than $ 1 $, so the largest of the numbers $ 1/a $, $ 1/b $, $ 1/c $ exceeds $ 1/3 $. Assuming $ a > b > c $, we conclude that $ c = 2 $. Equation (1) then takes the form $ 2a + 2b + ab = 2ab + 1 $, or $ (a - 2)(b - 2) = 3 $. Hence, $ a = 5 $, $ b = 3 $, $ c = 2 $. These numbers satisfy the conditions of the problem and are (up to permutation) its only solution.