As we have already seen phi²+ 1/phi² = 3. Now this can be represented in terms of the three sides of a right-angled triangle with phi² and 1/phi² representing the adjacent and opposite sides respectively and 3 the hypotenuse.

When the exponent is 3, phi³ - 1/phi³ = 4.
(However we cannot represent this result in two-dimensional geometrical
terms!)

However when we raise to the power of 4, we can represent the result in
terms of the right-angled triangle. Thus phi⁴ + 1/phi⁴ = 7.

Thus if one side of the triangle is 2.618034.. i.e. phi² and the other
.38197.. i.e. 1/phi² (squared) the hypotenuse will be the square root of 7.
Alternatively one side (2.618034..) = phi + 1 and the other (.38197..)
= 2 - phi.

So we can continue on in this manner and in the case of all even powers
represent the results in terms of the right-angled triangle.

Thus when we raise to the power of 6, phi⁶ + 1/phi⁶ = 18.

Thus if one side of the right-angled triangle is 4.236.. i.e. phi³ and the other .236.. i.e. 1/phi³, then the hypotenuse will be the square root of 18. This result is especially interesting as one side 4.236.. = (the square root of 5) + 2 and the other .236.. = (the square root of 5) - 2 .
Alternatively we can say that one side 4.236.. = 2phi + 1 and the other
.236.. = 2phi - 3.

Thus for all even powers we can express the opposite and adjacent sides of
the right-angled triangle in terms of simple combinations of phi (themselves
reflecting the Fibonacci sequence in a well-ordered fashion).

Thus when we raise to the power of 8 one side, phi⁴ = 3phi + 2.
The other 1/phi⁴ = 5 - 3phi and the hypotenuse will be the
square root of 47.
To the power of 10 one side of the Pythagorean triangle i.e. phi⁵ = 5phi + 3. The other 1/phi⁵ = 5phi - 8; the hypotenuse is then the square root of 123.
 

One remarkable connection - with a result on squares that I mentioned in a previous result - can be shown linking 123 to these expressions of phi.

123 = 89 + 34;

89 = 8² + 5²;
34 = 5² + 3²

Thus 123 = 5² + 3² + 8² + (- 5)² i.e. the squares of all the integers in our expressions for phi.

This result is not accidental and in fact universally holds.
So for example for the earlier result 47 = 3² + 2² + 5² + (- 3)²
.

When both opposite and adjacent sides represent phi⁶ and 1/phi⁶ respectively they can be expressed as 8phi + 5 and 13 - 8phi. (Notice how in the second case the sequence of these Fibonacci values
terms alternates with each new value!) The hypotenuse will then be the
square root of 322.
So 322 = 8² + 5² + 13² + (- 8) ² and of
course
322 = (8phi + 5)² + (13 - 8phi)². Yes, truly remarkable!

When we add the 4 numbers in our expression (without squares) the total = 18.

18² = 324 ( = 322 + 2).

This result is not accidental as the square (of the sum of the four numbers) in all cases = the original Lucas number + or - 2.

In fact the sign continually alternates. Thus in the earlier case where

123 = 5² + 3² + 8² + (- 5)²

the sum of terms (5 + 3 + 8 - 5) = 11;

and 11² = 123 - 2 .
 
 

(I would like to thank Mike Mc Dermott for his correspondence on this matter).