Almost a Square

Kim Seay

EMAT 6680

I begin this exploration with any square that has sides of integral length. I will start with the simple case of a square with a side length equal to 2cm.

The next step is to find the rectangle that has an area of one integer less than the original square (n^2 - 1), and also has sides of integral length.

Finding a Pattern

The challenge, is to find a pattern that can be applied to all such squares to find a rectangle that has an area of one integer less than the square. I explored a few more constructions before I was able to see the pattern.

As I began to see the pattern emerge, I made a table of squares with side lengths of 3cm - 20cm to test my theory.

 n n-1 n+1 (n-1)*(n+1) n^2 3 2 4 8 9 4 3 5 15 16 5 4 6 24 25 6 5 7 35 36 7 6 8 48 49 8 7 9 63 64 9 8 10 80 81 10 9 11 99 100 11 10 12 120 121 12 11 13 143 144 13 12 14 168 169 14 13 15 195 196 15 14 16 224 225 16 15 17 255 256 17 16 18 288 289 18 17 19 323 324 19 18 20 360 361 20 19 21 399 400

Now the pattern is clear to see. To find the rectangle with an area equal to one integer less than any square (with integral sides) simply use the sides n-1 and n+1. This will always work, because (n-1)(n+1) = n^2 +n -n - 1 = n^2-1.

Conclusion:

I like this exploration, because I have never thought about the difference of squares formula from this perspective. I memorized it over a decade ago, but never really pondered what it meant. This exploration made me do this, and I think it would do the same for students.

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