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Q&A

How can you "easily see that such squares [of side length $\sqrt{13}$ and $\sqrt{18}$] will not fit into the [4 × 4] grid"?

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Problem 2.4: How many squares of any size can be formed by connecting dots in the grid shown in Figure 2.2.

I skip p 31, but apprise me if you want me to include it.

  1. Side lengths of squares must be equal. Thus how can $m \neq n$ below?

  2. How do you most "easily see that such squares will not fit into the grid: there is no way to insert into the grid a square with side length $\sqrt(13)$ or with side length $\sqrt(18)$"? I can't easily see this. Perhaps I need an eye exam!

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David Patrick, BS Math & Computer Science, MS Math (Carnegie Mellon), PhD Math (MIT). Introduction to Counting & Probability (2005), pp 30, 32-3.

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Eye exam (1 comment)

1 answer

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The key point here is what $m$ and $n$ represent. Why is the side length of each square $\sqrt{m^2+n^2}$? Why do they have to be less than 4? And why does the author say that swapping $m$ and $n$ gives an identical case?

Because $m$ and $n$ are the legs of the right triangles that surround the square.

A square, placed on a grid, so that one of its edges spans m units vertically and n units horizontally. Dashed lines show that the same measurements are repeated on the square’s other edges.

(This seems like an important detail. Maybe it’s on page 31?)

Once we know this, we can “easily see” (author’s words) that the diagonal square is contained within a larger, axis-aligned square that has side length $m+n$, and so any combination where $m+n>4$ can’t fit on the grid.

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Actually for $\sqrt{18}$ it's even easier to see: Obviously $\sqrt{18} > \sqrt{16} = 4$, and clearly ... (2 comments)

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