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I have the following geometrical problem: let's assume a rectangle $a,b$, and we know that at the left top corner there is a square with area $l^2$. Is it possible to fill the remaining whith pure squares? It is obvious that those squares cannot be of the same size, since $b\neq a$. Many thanks for your help.

pikunsia
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    Your last statement is false. It's not at all obvious that those squares cannot be of the same size. Consider a rectangle of size 2x3 with a square of size 1x1 in the corner. You can fill the remaining surface with 5 squares. – Stef Feb 12 '23 at 14:30
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    Please [edit] the question to show us what you tried. Have you examined a few cases where you know the answer is definitely yes or no? Show us those. – Ethan Bolker Feb 12 '23 at 14:31
  • Remark: A rectangle is (x, y)-symmetric. It means that the position information of $l^2$ vertex position is irrelevant. Also, it is important to remark that l \leq \min(a, b). – Bruno Lobo Feb 12 '23 at 14:32
  • As @Stef has noted, you can do it sometimes, so the question is when. The remainder of the square is made up of 2 rectangles. When can you fill a rectangle exactly with finitely many squares? You can always do it with infinitely many squares. – Peter Feb 12 '23 at 14:37
  • By a result of Max Dehn in 1903, this is possible iff $\frac{a}{\ell}, \frac{b}{\ell} \in \mathbb{Q} \cap [1,\infty)$. see this answer for a related question which contains enough stuff to justify this claim. – achille hui Feb 12 '23 at 15:27

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As far as I can see, there is:

Statement: A rectangle with $(a, b)$-sides has a inscribed $l$-sided square, for condition $l \leq \min(a, b)$, with coincidental vertex to its circumscribed rectangle.

Remark 1: The square breaks the rectangle in quadrants. Any solution that fills these quadrants is a feasible solution.

Remark 2: You can feel the solution with identical $\varepsilon$-sided squares such that the remaining $(a b - l^2)$-area is filled. It means, the amount of $\varepsilon$-squares is $\frac{ab - l^2}{\varepsilon^2}$. You must noticed that this rational fraction must be a natural number. This means, there are only a subset of possible solutions among $0 < \varepsilon < \min(l-a, \, l-b)$ such that equality $\frac{ab - l^2}{\varepsilon^2} = n$ hold, for some natural $n \in \mathbb{N}_{>0}$.

Bruno Lobo
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