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: $$0 , 1 , 1 , 2 , 3
, 5 , 8, 13 , 21 , 34 , \cdots$$so that $$F_0 =0, F_1 =
1, F_2 = 1, F_3=2, F_4 = 3, \cdots$$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.
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
$F_{2n}$ (note this will be a square with a square cut out of one
corner) and $F_{2n+1}$ ( 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 $F_{21}$ .