Henry Ernest Dudeney/Puzzles and Curious Problems/129 - Squares and Cubes/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $129$

Squares and Cubes
Find two different numbers such that the sum of their squares shall equal a cube, and the sum of their cubes equals a square.


Solution

We have:

$625^2 + 1250^2 = 125^3$

while:

$625^3 + 1250^3 = 46 \, 875^2$


Proof

We wish to find integral solutions to:

\(\text {(1)}: \quad\) \(\ds x^2 + y^2\) \(=\) \(\ds z^3\)
\(\text {(2)}: \quad\) \(\ds x^3 + y^3\) \(=\) \(\ds w^2\)

Dudeney states that if one number is $625 m^6$ and the other number double the first, you can get any number of solutions of a particular series.

The answer given is for $m = 1$.

He probably exploited the fact that $1 + 2^3 = 3^2$.

Therefore for any $k$:

$k^3 + \paren {2 k}^3 = 9 k^3$

$9 k^3$ would be a square if $k$ is itself a square.

So we write $k = n^2$:

$\paren {n^2}^3 + \paren {2 n^2}^3 = \paren {3 n^3}^2$

and we see that $(2)$ is satisfied with $x = n^2, y = 2 n^2, w = 3 n^3$.


Substituting this to $(1)$, we have:

$\paren {n^2}^2 + \paren {2 n^2}^2 = 5 n^4 = z^3$

Since $5 n^4$ is a cube, $n$ must be divisible by $5$.

By writing $n = 5 a$, we still require $5^5 a^4$ to be a cube.

Hence $a$ must again be divisible by $5$.

By writing $a = 5 b$, we only require $5^9 b^4$ to be a cube.

Thus $b$ must itself be a cube.

Finally we write $b = m^3$.

$k = n^2 = 25 a^2 = 625 b^2 = 625 m^6$

and we see that:

$\paren {625 m^6}^2 + \paren {1250 m^6}^2 = 5^9 m^{12} = \paren {125 m^4}^3$

which is a cube, so $(1)$ is also satisfied.


Substituting $x = 625 m^6, y = 1250 m^6$ back to $(2)$, we have:

$\paren {625 m^6}^3 + \paren {1250 m^6}^3 = 2 \, 197 \, 265 \, 625 m^{18} = \paren {46 \, 875 m^9}^2$

and thus we have obtained a general solution.

$\blacksquare$


Sources

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