Dissection puzzle for area 49 to area 50

Question

49 and 50 are close, as are 288 and 289. That allows a grid illusion. If cut out of wood, perhaps with coloring on the border as an "assistance", the pieces could be dumped out of the tray, flipping them, then scrambled a bit, and the solver could be asked to put them back into the tray. But there would be a gap. "No, that's not right." The owner would dump the pieces out again, then place them in perfectly, then dump them again, and invite the solver to try again.

Are there better dissections that can be used, which stay on the grid lines? Ideally, I'd like to stay under 7 pieces, of roughly the same area, and I'd like the hole in the second figure to be square.

Joseph Kisenwether sent me the following, which is pretty good.

I'm not looking for old versions of the Missing Square Puzzle. I'm looking for new dissections specifically for squares with area 49 and 50, or areas 288 and 289, and where all dissection lines are in the directions of queen moves.

Answer 1

Have you ever seen the Missing Square Puzzle?

Maybe this isn't exactly what you're looking for, but it is along the same lines. At the bottom of the wikipedia article, you'll find similar puzzles (like Sam Lloyd's paradoxical disection), which also may help. Here is one that is fairly interesting.

Answer 2

I guess the following might be also in the Wikipedia article Missing square puzzle, but A. Beutelspacher explained this very well in his book "Diskrete Mathematik für Einsteiger" (ISBN 978-3-8348-1248-3)

I am not sure if this is what you're looking for, but it might still be interesting.

Fibonacci numbers

Fibonacci numbers are integers that are defined as

$$f_0 := 0, f_1 := 1, \forall n \in \mathbb{N}_{\geq 2}:f_n := f_{n-1} + f_{n-2}$$

The first fibonacci-numbers are

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

Simpson-Identity

The Simpson-Identity states that

$$\forall n \in \mathbb{N}_{\geq 2}: f_{n+1} \cdot f_{n-1} - f_n^2 = (-1)^n $$

Proof: with induction

Missing square puzzle

(I've added this as SVG to Wikipedia, just in case you want to see this bigger: Missing-squre-fibonacci.svg)

So, whats going on here?

The square on the left has a size of $f_n^2$, the rectangle on the right seems to have a size of $f_{n-1} \cdot f_{n+1}$ which makes (according to Simpson's identity) a difference of $\pm 1$. Now you can make $n$ as big as you want, the difference will still be $\pm 1$. This means, you can make the difference as difficult to see as you want.