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 A1 = x + 1/x = 1 and note that x has no real solutions. Let
An = xn+ 1
xn

.

We have A0 = 1 + 1 = 2.

For
A2 = x2+ 1
x2

, as
(x + 1/x)2 = x2 + 2 + 1
x2
 = 1

so A2 = A1 - A0 = 1 - 2 = -1.

For
A3 = x3+ 1
x3

, since
A2 = x2 + 1
x2
 = (x2 + 1
x2
 )(x + 1/x) = x3 + x + 1/x + 1
x3
 = A3 + A1

, therefore A3 = A2 - A1 = -1 -1 = -2.

In general
An-1 = xn-1 + 1
xn-1
 = (xn-1 + 1
xn-1
 )(x + 1/x) = xn + xn-2 + 1
xn
 + 1
xn-2
 = An + An-2

, therefore An = An-1 - An-2.

From A0=2 and A1=1 we can generate the whole sequence of An 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:

x3 = -1 and hence that x4 = -x, x5 = -x2 and x6 = -x3 = 1. If x + 1/x=1 then x2-x + 1 = 0 and x ¹ -1 so (x+1)(x2 - x +1)=x3 +1 = 0. Hence x is the cube root of -1 so x4 = -x and x7 = x. From this it is easy to show that
xn + 1
xn

takes the values 1, -1, -2, -1, 1, 2, 1, -1, .... cyclically.