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
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
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).