A set of n positive number, x1, x2, ...xn, n > = 3
are defined as:

x1=1+(1/x2)
x2=1+(1/x3), etc.
also, xn=1+(1/x1)

Could someone please give me a hint on how to prove that x1=x2=...xn. If this is the case, then x1 is the golden ratio, but I can't see how to prove it (which is annoying,because it seems like it should be something really obvious). For any value of x1 other than the golden ratio, the sequence appears to diverge on (though never reach) the golden ratio, meaning that the series can't be "circular" (not sure what the proper term is). However, I have no idea how to prove this, and this method seems too complicated to be the answer to a question such as this. I also have a formula for xj, but I can't prove this.
My formula is:

x_j = -1^j+1 S_j x₁ + (-1)^j S_j +1

actually, I'm not going to be able to type this out, but basically, I noticed that when expressing x1, x2, etc. in terms of x1 the numerator and denominator followed a slightly adapted version of the Fibonnaci sequence (1,0,1,2,3,5,etc.).

Please could you give me some hints on what kinds of methods to use. Thanks

Obviously 1,0,1,2,3,5, etc. should actually be 1,0,1,1,2,3,5,etc.

Show that x_i -x_i+1 = -x_i x_i+1 (x_i-1 -x_i ) (where the subscripts are interpreted cyclically if necessary). You can deduce that for any i we have (x_i -x_i+1 ) = (-1)ⁿ x₁ ² ...x_n ² (x_i -x_i+1 ) and the result follows.

a way of proving x1 = x2 = ... = xn is to assume xi> x(i+1)
=> 1+1/x(i+1) > 1+1/x(i+2)
=> 1/x(i+1) > 1/x(i+2)
=> x(i+1) less than x(i+2)
=> absurdity

now repeat for xi less than x(i+1)

which all => xi=(xi+1)

Alex