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. It is now easy to show the values of

xn +

1 xn

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

Method 2

(x+ 1 x )(xn + 1 xn ) = xn+1 + 1 xn+1 + xn-1 + 1 xn-1 .

But x+¹/_x=1 so

xn+1 + 1 xn+1 = æ ç è xn + 1 xn ö ÷ ø - æ ç è xn-1 + 1 xn-1 ö ÷ ø .

Now

x1+

1 x1

=1

and

x0 +

1 x0

=2

. Hence

x2+ 1 x2 =1 - 2 = -1

x3+ 1 x3 = -1 - 1 = -2

x4+ 1 x4 = -2 - (-1) = -1

x5+ 1 x5 =-1 - (-2) = 1

x6+ 1 x6 =1 - (-1) = 2

x7+ 1 x7 =2 - 1 = 1

So the process repeats itself in a 6-cycle.

Method 3

Write x = e^logx = e^y and ny = logxⁿ. Then x + ¹/_x=1 is equivalent to 2coshy = 1 and

xn +

1 xn

= 2 coshny

. Now

sinh2 y =

-3 4

so

coshy + sinhy = ey =

1-iÖ3 2

. So e^y is at

(1, -

p 3

)

in the Argand diagram.

So all powers of e^y lie on the vertices of a regular hexagon, centre the origin and hence e^6y=1. So

xn +

1 xn

takes values in a 6-cycle.

See Mathematical Gazette December 1969 Number 386