The golden ratio number φ = (1+√5)/2 possesses many interesting properties.

inter alia
For any even integer n: φⁿ + 1 /φⁿ is an integer
For any odd integer n: φⁿ - 1 /φⁿ is an integer

Prove the above.

Outline solution

Using phi² = phi + 1 repeatedly to break down phiⁿ to a

linear function of phi shows that   phiⁿ = F_n*phi + F_n-1 ,

where F_i are Fibonacci numbers (F₁ = 1, F₂ = 1, F₃ = 2 …).

(for example, phi⁶ = F₆*phi + F₅ = 8*phi + 5)

This can be extended to negative values of n by using the

bidirectional Fibonacci series:  …5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5…

Note that for even n,  F_n = -F_-n and for odd n, F_n = F_-n .

Thus     phiⁿ + 1/phiⁿ      = (F_n*phi + F_n-1)+ (F_-n*phi + F_-n-1)

                                    = (F_n + F_-n)*phi + F_n-1 + F_-n-1      

so for even n the bracket is zero and the result is an integer.

Also       phiⁿ – 1/phiⁿ     = (F_n*phi + F_-n) – (F_-n*phi + F_-n-1)

                                    = (F_n – F_-n)*phi + F_-n – F_-n-1

so for odd n the bracket is zero and the result is an integer.

_________

PS Is there a simple way to produce greek letters in these

postings?  I’m pasting Word text.

Posted by Harry on 2016-10-18 18:44:16