Henry Ernest Dudeney/Modern Puzzles/95 - Squares and Cubes/Solution

by : $95$

 * Squares and Cubes

Solution
We have:

Proof
We show this solution is minimal by showing that the larger number is at least $10$.

We look at the first equation:
 * $x^2 - y^2 = a^3 = \paren {x - y} \paren {x + y}$

Since $y^2 > 0$ and $5^3 > 10^2$, we only consider $a = 1, 2, 3, 4$.

$1$ cannot be the difference of two squares.

Since $2^3 = 1 \times 8 = 2 \times 4$, we check:
 * $\tuple {x, y} = \tuple {3, 1}$

as the other pair are not integers.

This gives:
 * $3^3 - 1^3 = 26$

which is not a square.

Since $3^3 = 1 \times 27 = 3 \times 9$, we check:
 * $\tuple {x, y} = \tuple {6, 3}$

as the other pair has $x > 10$.

This gives:
 * $6^3 - 3^3 = 189$

which is not a square.

Since $4^3 = 1 \times 64 = 2 \times 32 = 4 \times 16 = 8 \times 8$, we check:
 * $\tuple {x, y} = \tuple {10, 6}$

as the other pairs either has $x > 10$ or $y = 0$.

This gives:
 * $10^3 - 6^3 = 784 = 28^2$

which is our smallest solution.