Look first at the Gnomon 1 problem.
To investigate further the representation of Fibonacci numbers by gnomons it will help to use a little algebra. We denote by

F_{r}

the

r^{th}

Fibonacci number in the sequence:

$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \dots$

so that $F_{0} = 0, F_{1} = 1, F_{2} = 1, F_{3} = 2, F_{4} = 3,$ etc.

Now the Fibonacci rule can be written as

$F_{r + 1} = F_{r} + F_{r - 1}$

Here are the gnomons for $F_{4}, F_{5}, F_{6}$ and $F_{7}$ . Note that each gnomon has six sides (hexagonal) and that when $r$ is even the $F_{r}$ -gnomon is a square with a square cut out of the corner and when $r$ is odd the $F_{r}$ -gnomon is a rectangle (not a square) with a rectangle cut out of the corner.

F4 gnomon. F5 gnomon. F6 gnomon. F7 gnomon.

Draw some gnomons for yourself and mark the six lengths of the edges of the gnomons with the corresponding Fibonacci numbers. Draw dotted lines dividing each gnomon into two parts illustrating the Fibonacci rule. [You may like to cut out a set of gnomons for yourself so that you can fit them together, two by two, like pieces of a jigsaw to demonstrate the Fibonacci rule.]

To give the solution to this question, draw the gnomons for $F_{2 n}$ (note this will be a square with a square cut out of one corner) and $F_{2 n + 1}$ ( two squares attached to each other making a rectangle with a rectangle cut out of the corner). In each case label the lengths of the six sides with the Fibonacci numbers as $F_{n - 1}, F_{n}, F_{n + 1},$ etc. thus giving a general specification for the gnomon corresponding to any Fibonacci number. Now check that your specification works for the Fibonacci numbers $F_{20}$ and $F_{21}$ .