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'Sextet' printed from https://nrich.maths.org/
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 $A_1 = x + {1\over x} = 1$ and note that $x$ has no real
solutions. Let $A_n = x^n+ {1\over x^n}$.
We have $A_0 = 1 + 1 = 2$.
For $A_2 = x^2+ {1\over x^2}$, as $(x + {1\over x})^2 = x^2 + 2 +
{1\over x^2}= 1$ so $A_2 = A_1 - A_0 = 1 - 2 = -1$.
For $A_3 = x^3+ {1\over x^3}$, since $A_2 = x^2 + {1\over x^2} =
(x^2 + {1\over x^2})(x + {1\over x}) = x^3 + x + {1\over x} +
{1\over x^3} = A_3 + A_1$, therefore $A_3 = A_2 - A_1 = -1 -1 =
-2$.
In general $A_{n-1} = x^{n-1} + {1\over x^{n-1}} = (x^{n-1} +
{1\over x^{n-1}})(x + {1\over x}) = x^n + x^{n-2} + {1\over x^n} +
{1\over x^{n-2}} = A_n + A_{n-2}$, 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 $x + {1\over x}=1$ 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 $x^n
+ {1\over x^n}$ takes the values 1, -1, -2, -1, 1, 2, 1, -1, ....
cyclically.