3.15 Problem 15: One Unit Missing

We take a smaller instance for the same problem (see Figure at the right). Now one can see immediately where the problem is: the diagonal in the 3 × 8 rectangle ist not a straight line. Hence, none of the forms is a triangle. A quick calculation also shows that they cannot be triangles. The proportion of the height and the length of these triangles is 3 8. At the position 5 at the length, the height of this “triangle” is 2. But we know that 3 8 = 2 5. Hence, these forms are not triangles.

   

PIC

How did we discover these figures? They are “special numbers” 3 × 8 is almost 5 × 5, and 8 × 21 is almost 13 × 13. Sounds familiar? 1, 1, 2, 3, 5, 8, 13, 21, 34,... These are consecutive Fibonacci numbers. These numbers are defined by:

F1 =  1 ,  F2 = 1 ,  Fn =  Fn-1 + Fn- 2 ,  with n ≥  2

The Fibonacci numbers have many interesting properties, one of them is

Fn+1 ⋅ Fn -1 - F2 = (- 1)n ,  with n ≥ 1
               n

Proof (by induction):

  1. The property is true for n = 1, since 1 0 - 12 = (-1)1.

  2. Supposing the property is valid for n, then it is valid for n + 1, namely we substitute Fn-1 with Fn+1 -Fn in the property and we get:

      2                2       n
Fn+1 - Fn+1Fn  - F n = (- 1)

    By multiplying with -1 and transforming it we finally get:

                 2         n+1
Fn+2 ⋅ Fn - F n+1 = (- 1 )

    That proves the property.

  3. In particular, for n = 6 we have: 13 5 - 82 = (-1)6 producing the initial graph of the problem. We also have: 8 3 - 52 = (-1)5.