Thank you Andrei Lazanu, age 12, School No. 205 Bucharest, Romania for this solution.
I drew separately the gnomons for F r with r even and r odd, up to r = 12. I replaced the lengths of the sides with the terms in the Fibonacci sequence, and then I tried to find a rule for the gnomons corresponding to F 2n and F 2n+1 respectively.
For Fr with r even, I obtained the following table, with l j for j = 1, 2,..., 6, the lengths of the sides of the gnomon:
| l 1 | l 2 | l 3 | l 4 | l 5 | l 6 | |
| F 2 | F0 | F 0 | F 1 | F 1 | F 2 | F 2 |
| F 4 | F 4 | F 4 | F 2 | F 2 | F 3 | F 3 |
| F 6 | F 2 | F 2 | F 3 | F 3 | F 4 | F 4 |
| F 8 | F 3 | F 3 | F 4 | F 4 | F 5 | F 5 |
| F 10 | F 4 | F 4 | F 5 | F 5 | F 6 | F 6 |
| F 12 | F 5 | F 5 | F 6 | F 6 | F 7 | F 7 |
So, for the gnomon F2n I observed the following rule:
| F 2n | F n-1 | F n-1 | F n | F n | F n+1 | F n+1 |
This is an inductive process and at the end I have to verify it.
For F r with r odd, I obtained the following table, with lj for j = 1, 2,..., 6, the lengths of the sides of the gnomon:
| l 1 | l 2 | l 3 | l 4 | l 5 | l 6 | |
| F 5 | F 4 | F 3 | F 3 | F 2 | F 2 | F 1 |
| F 7 | F 5 | F 4 | F 4 | F 3 | F 3 | F 2 |
| F 9 | F 6 | F 5 | F 5 | F 4 | F 4 | F 3 |
| F 11 | F 7 | F 6 | F 6 | F 5 | F 5 | F 4 |
For the gnomon F2n+1 the following rule could be observed:
| F 2n+1 | F n+2 | F n+1 | F n+1 | F n | F n | F n-1 |
Using these rules, the gnomons for F20 and F21 have the following lengths of the sides:
| F 20 | F 9 | F 9 | F 10 | F 10 | F 11 | F 11 |
| F 21 | F 12 | F 11 | F 11 | F 10 | F 10 | F 9 |
Keeping into account that the area of the gnomon equals the corresponding number in the Fibonacci sequence, and also the shape of each kind of the gnomon, I found the following recursive relations:
F 2n = F n (F n-1 + F n+1) F 2n+1 = F n+2 F n + F n-1 F n+1
I hope that both relations are satisfied, but I don't know so much algebra to verify them.
As discovered in the solution to Gnomon I from May 2001 we also have the relations
F 2n = (F n+1 ) 2 - (F n-1 ) 2
and
F 2n+1 =(F n+1 ) 2 + (F n ) 2 .
These four relations are illustrated in the gnomon diagram below.
The diagram shows gnomons whose areas (labelled in black) are Fibonacci numbers. It shows the lighter and darker yellow gnomons F2n-1 and F2n fitting together to make F 2n+1 with the total yellow area given by
F 2n-1 +F 2n = F 2n+1 .
The yellow gnomon representing F 2n+1 fits together with the blue gnomon representing F2n to make the gnomon for F
2n+2.
The red labelling shows the lengths of the edges.