I noticed in my mathematical meanderings involving Phi I often had to reduce algebraic formulae involving powers of Phi. I began to wonder if there was an elegant way to represent those powers that is both useful and simple. I first began with the positive integer powers, 1, 2, 3, ….

f^{0} = 1

f^{1} = ½[Ö
5 + 1]

f^{2} = ¼[2Ö
5 + 6] = ½[Ö
5 + 3]

f^{3} = ¼[4Ö
5 + 8] = ½[2Ö
5 + 4]

f^{4} = ½[3Ö
5 + 7]

f^{5} = ¼[10Ö
5 + 22] = ½[5Ö
5 + 11]

f^{6} = ¼[16Ö
5 + 36] = ½[8Ö
5 + 18]

f^{7} = ½[13Ö
5 + 29]

f^{8} = ¼[42Ö
5 + 94] = ½[21Ö
5 + 47]

f^{9} = ¼[68Ö
5 + 152] = ½[34Ö
5 + 76]

And so I noticed the Fibonacci series as the first coefficient of the radical easily enough, and I speculated about the second term for a short bit, quickly verifying that it is indeed consecutive elements in the Lucas series. Thus, a nice way to represent the general form for all non-negative integer powers of Phi yields:

f^{n} = ½[F_{(n)Ö
}5 + L_{(n)}], n ³
0 (1.1)

Nice! Further analysis of the negative powers followed thus:

f^{-1} = f - 1 = ½[Ö
5 - 1]

f^{-2} = ½[3 - Ö
5]

f^{-3} = ¼[4Ö
5 - 8] = ½[2Ö
5 - 4]

f^{-4} = ½[7 - 3Ö
5]

f^{-5} = ½[5Ö
5 - 11]

The pattern here was quickly becoming apparent, resulting in the following for all non-positive integers n:

f^{n} = ½[(-1)^{n – 1}F_{(|n|)Ö
}5 + (-1)^{n} L_{(|n|)}], n £
0 (1.2)

Combining the two resulted in the following relation over all integers n:

f^{n} = ½[(-1)^{(a)}F_{(|n|)Ö
}5 + (-1)^{(b)} L_{(|n|)}] (1.3)

where:

n |
even and negative |
odd and negative |
0 |
even and positive |
odd and positive |

a |
1 |
0 |
0 |
0 |
0 |

b |
0 |
1 |
0 |
0 |
0 |

Or, a = n – 1, b = n when n is negative, a = b = 0 all other n.

In all of the above formulae F_{(n) }represents the n^{th} member of the Fibonacci series, where F_{(0)} = 0, F_{(1)} = 1, and F_{(n+2)} = F_{(n)} + F_{(n + 1)}. Likewise L_{(n)} denotes the n^{th} member of the Lucas series, where L_{(0)} = 2, L_{(1)} = 1, and L_{(n+2)} = L_{(n)} + L_{(n + 1)}.

Some interesting twists on the above give the following Fibonacci and Lucas relationships:

F_{(n) }= (2f^{n} - L_{(n)})/Ö
5 (1.4)

and

L_{(n) }= 2f^{n} - F_{(n)Ö
}5 (1.5)

I have not seen these five relations published, either on the Internet or on paper, having read a dozen or so books on the subject. I have not gotten hold of S. Vajda’s "*Fibonacci and Lucas Numbers, and the Golden Section*", Ellis Horwood Limited, Chichester, 1989. I have a feeling they may be printed there, but until such time as I can verify, as the publication is currently out of print, I hope these will stand as my contributions.