Solution to: Square Slicing

Assume that the three square boards have sides of p, q, and r cm, arranged from smallest to largest. The board with side p produces (p − 2)² unvarnished cubes. The rest of the total p² cubes are partially varnished. This means that the number of partially varnished cubes is equal to p² - (p − 2)², which can be simplified to:

p² − (p² − 4p + 4) = 4p − 4.

This also applies to the other two boards. Therefore, we need to find a solution for the following equation:

(p − 2)² + (q − 2)² + (r − 2)² = (4p − 4) + (4q − 4) + (4r − 4).

We can rewrite this as: (p − 4)² + (q − 4)² + (r − 4)² = 24.

Since p, q, and r are integers, this means that the sum of three whole squares must equal 24. The only whole squares under 24 are 0, 1, 4, 9, and 16. There is only one way to achieve the sum of 24 with three of these squares: 4 + 4 + 16 = 24. Because pp, q, and r are different, this means that these squares must be formed as follows: (-2)² + 2² + 4² = 24.

From this, it follows that (p - 4) = -2, (q - 4) = 2, and (r - 4) = 4. This leads to the values p = 2, q = 6 en r = 8.

Thus, the sides of the boards were 2, 6, and 8 cm.

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