The history of mathematics is an abundant field for making surprising discoveries. Some luck is necessary, but whereas the literary tradition has been extensively explored by scholars, finding a nugget in the history of mathematics is not so difficult as turning up a new fragment by Jane Austen, or a lost piece by Mozart. In my case I struck lucky in the theory of matrices.

As is well known, the theory of matrices arose during the 1850s – a useful benchmark is the publication of Arthur Cayley's celebrated memoir of 1858, A memoir on the theory of matrices, a paper which generations of mathematicians have taken as signalling the beginning of matrix theory. Even today the opening paragraphs serve as a useful introduction to the subject.

In maths lessons in secondary school, on certain occasions my teacher entertained our class with the following game. Each player starts with an empty 2 × 5 grid, as shown in Figure 1(a). In each round of the game, a die is rolled, and each player has to choose one of the empty boxes in their grid and fill it with the number rolled. For example, after 3 rounds, a player's grid may look like the one in Figure 1(b). After 10 rounds everyone's grid is filled, forming two 5-digit numbers. A player's score is the sum of her/his two 5-digit numbers. Players with the largest sum win the game. We shall call this the ‘dice-and-numbers game’. It is easy to think of other variants. Any child who is comfortable with column addition and understands how place value affects the size of numbers will develop a reasonable strategy for the game.

The Lyness equation (1)

\begin{equation}{X_{n + 1}} = \frac{{{X_n} + a}}{{{X_{n - 1}}}},\,(a,{x_1},{x_2} > 0)\end{equation}

The well-known Fibonacci numbers Fn are defined by the recurrence relation

Fn = Fn – 1 + Fn – 2. (1)

together with the starting values F0 = 0, F1 = 1, or equivalently F1 = F2 = 1.

We record the first few:

The recurrence relation can also be applied backwards in the form Fn = Fn + 2 – Fn + 1 to define Fn for n < 0. An easy induction verifies that F−n = (−1)n – 1Fn for n > 0.