Lucas numbers relation to Φ

Question

So, the Lucas numbers are 2,1,3,4,7,11... Let L(n) be nth lucas number

Fibonacci numbers are 1,1,2,3,5,8,13,21...

Φ^n=F(n+1)+F(n-1), F=Fibonacci number and n=nth

So, if I say n=5, then Φ^n=F(6)+F(4)=11 (8+3).

11=6th Lucas number.

For n=10, Φ^n=F(11)+F(9)=89+34=123 =L(11)

Again, Φ^n=L(n+1)

So, Φ^n=F(n+1)+F(n-1)=L(n+1). How is L(n)=Φ^n?

Answer 1

You can easily show that ø+1/ø = sqrt(5), and by beniot's formula, you have Lucas = Fibonacci * sqrt(5).

Using base ø, such that 11 = 100, you can show that 10.1 square = 5, as

10.1² =  102.01  = 13.01 = 4.11 = 5.