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

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Modern Puzzles by Henry Ernest Dudeney: $95$

Squares and Cubes
Can you find two whole numbers, such that the difference of their squares is a cube and the difference of their cubes is a square?
What is the answer in the smallest possible numbers?


Solution

We have:

\(\, \ds 10^2 - 6^2 = \, \) \(\ds 100 - 36\) \(=\) \(\ds 64\) \(\ds = 4^3\)
\(\, \ds 10^3 - 6^3 = \, \) \(\ds 1000 - 216\) \(=\) \(\ds 784\) \(\ds = 28^2\)


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.

$\blacksquare$


Sources

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