The golden ratio number φ = (1+√5)/2 possesses many interesting properties.
inter alia
For any even integer n:
φ^{n} + 1 /φ^{n} is an integer
For any odd integer n:
φ^{n}  1 /φ^{n} is an integer
Prove the above.
Outline
solution
Using phi^{2} = phi + 1 repeatedly to break down phi^{n} to a
linear function of phi shows that phi^{n}
= F_{n}*phi + F_{n1} ,
where F_{i} are Fibonacci numbers (F_{1} = 1, F_{2 }=
1, F_{3} = 2 …).
(for example, phi^{6} = F_{6}*phi + F_{5} = 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 phi^{n}
+ 1/phi^{n} = (F_{n*}phi + F_{n1})+ (F_{n}*phi
+ F_{n1})
=
(F_{n} + F_{n})*phi + F_{n1} + F_{n1}
so for even n the bracket is zero and the result is an integer.
Also phi^{n} – 1/phi^{n} = (F_{n}*phi
+ F_{n}) – (F_{n}*phi + F_{n1})
= (F_{n}
– F_{n})*phi + F_{n} – F_{n1}
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 20161018 18:44:16 