Oliver from Olchfa School and Simon from Elizabeth College, Guernsey both proved a general property of the sequence, namely that each term is the difference of the two previous terms. From this they found that the sequence is a repeating cycle of six values. You may like to consider more general sequences $a, b, b - a, . . .$ with the property that each term is the difference of the two previous terms and investigate whether such sequences are always cyclical.

This is Oliver's solution:

Let $A_{1} = x + \frac{1}{x} = 1$ and note that $x$ has no real solutions. Let $A_{n} = x^{n} + \frac{1}{x^{n}}$ .

We have $A_{0} = 1 + 1 = 2$ .

For $A_{2} = x^{2} + \frac{1}{x^{2}}$ , as $(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x^{2}} = 1$ so $A_{2} = A_{1} - A_{0} = 1 - 2 = - 1$ .

For $A_{3} = x^{3} + \frac{1}{x^{3}}$ , since $A_{2} = x^{2} + \frac{1}{x^{2}} = (x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x}) = x^{3} + x + \frac{1}{x} + \frac{1}{x^{3}} = A_{3} + A_{1}$ , therefore $A_{3} = A_{2} - A_{1} = - 1 - 1 = - 2$ .

In general $A_{n - 1} = x^{n - 1} + \frac{1}{x^{n - 1}} = (x^{n - 1} + \frac{1}{x^{n - 1}}) (x + \frac{1}{x}) = x^{n} + x^{n - 2} + \frac{1}{x^{n}} + \frac{1}{x^{n - 2}} = A_{n} + A_{n - 2}$ , therefore $A_{n} = A_{n - 1} - A_{n - 2}$ .

From $A_{0} = 2$ and $A_{1} = 1$ we can generate the whole sequence of $A_{n}$ as follows: 2, 1, -1, -2, -1, 1, 2, 1, -1, ... We can see that the sequence is a repeating pattern of 2, 1, -1, -2, -1, 1 for successive values of $n$ with a period of 6.

Rupert from Wales High School noted that:

$x^{3} = - 1$ and hence that $x^{4} = - x$ , $x^{5} = - x^{2}$ and $x^{6} = - x^{3} = 1$ . If $x + \frac{1}{x} = 1$ then $x^{2} - x + 1 = 0$ and $x \neq - 1$ so $(x + 1) (x^{2} - x + 1) = x^{3} + 1 = 0$ . Hence $x$ is the cube root of $- 1$ so $x^{4} = - x$ and $x^{7} = x$ . From this it is easy to show that $x^{n} + \frac{1}{x^{n}}$ takes the values 1, -1, -2, -1, 1, 2, 1, -1, .... cyclically.