As we have already seenphi. Now this can be represented in terms of the three sides of a right-angled triangle with phi^{2}+ 1/phi^{2}= 3^{2}and 1/phi^{2}representing the adjacent and opposite sides respectively and 3 the hypotenuse.When the exponent is

3,phi.^{3}- 1/phi^{3}= 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. Thusphi^{4}+ 1/phi^{4}= 7.Thus if one side of the triangle is

2.618034.. i.e.phiand the other^{2}.38197..i.e.1/phi(squared) the hypotenuse will be the square root of^{2}7.

Alternatively one side(2.618034..)= phi + 1and 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,phi+^{6}1/phi^{6}= 18.Thus if one side of the right-angled triangle is

4.236..i.e. phiand the other^{3}.236..i.e.1/phi, then the hypotenuse will be^{3}the square root of 18. This result is especially interesting as one side4.236.. = (the square root of 5) + 2and the other.236.. = (the square root of 5) - 2.

Alternatively we can say that one side4.236..= 2phi + 1and 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

8one side,phi^{4}= 3phi + 2.

The other1/phi^{4}= 5 - 3phiand the hypotenuse will be the

square root of47.

To the power of10one side of the Pythagorean triangle i.e.phi^{5}= 5phi + 3. The other1/phi; the hypotenuse is then the square root of^{5}= 5phi - 8123.

One remarkable connection - with a result on squares that I mentioned in a previous result - can be shown linking

123to these expressions ofphi.

123 = 89 + 34;

89 = 8^{2}+ 5;^{2}34 = 5+^{2}3^{2}^{}Thus

123 = 5^{2 }+ 3^{2}+ 8+ (^{2}- 5)i.e. the squares of all the integers in our expressions for phi.^{2}This result is not accidental and in fact universally holds.

So for example for the earlier result47 = 3^{2}+ 2^{2}+ 5^{2 }+ (- 3)^{2}

.When both opposite and adjacent sides represent

phiand^{6}1/phirespectively they can be expressed as^{6}8phi + 5and13 - 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 of322.

So322 = 8^{2}+ 5and of^{2}+ 13^{2}+ (- 8)^{ 2}

course322 = (8phi + 5)+^{2}(13 - 8phi). Yes, truly remarkable!^{2}When we add the 4 numbers in our expression (without squares) the total

= 18.

18^{2 }= 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 = 5^{2 }+ 3^{2}+ 8^{2}+ (- 5)^{2}the sum of terms

(5 + 3 + 8 - 5) =11;and

11^{2 }= 123 - 2 .

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