(related to Problem: The Square of Veneer)

Any square number may be expressed as the sum of two squares in an infinite number of different ways. The solution of the present puzzle forms a simple demonstration of this rule. It is a condition that we give actual dimensions.

In this puzzle, I ignore the known dimensions of our square and work on the assumption that it is $13n$ by $13n.$ The value of n we can afterwards determine. Divide the square as shown (where the dotted lines indicate the original markings) into $169$ squares. As $169$ is the sum of the two squares $144$ and $25,$ we will proceed to divide the veneer into two squares, measuring respectively $12\times12$ and $5\times5;$ and as we know that two squares may be formed from one square by dissection in four pieces, we seek a solution in this number. The dark lines in the diagram show where the cuts are to be made. The square $5\times5$ is cut out whole, and the larger square is formed from the remaining three pieces, $B, C,$ and $D,$ which the reader can easily fit together.


Now, $n$ is clearly $\frac{5}{13}$ of an inch. Consequently, our larger square must be $\frac{60}{13} \times \frac{60}{13}$ in., and our smaller square $\frac{25}{13} \times \frac{25}{13}$ in. The square of $\frac{60}{13}$ added to the square of $\frac{25}{13}$ is 25. The square is thus divided into as few as four pieces that form two squares of known dimensions, and all the sixteen nails are avoided.

Here is a general formula for finding two squares whose sum shall equal a given square, say $a^2.$ In the case of the solution of our puzzle $p = 3,$ $q = 2,$ and $a = 5$:

$$\frac{2pqa}{p^2+q^2}=x,\quad\quad\frac{\sqrt{a^2( p^2 + q^2)^2 - (2pqa)^2}}{p^2 + q^2} =y.$$

Here $x^2 + y^2 = a^2.$

Thank you to the contributors under CC BY-SA 4.0!



Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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