Henry Ernest Dudeney/Puzzles and Curious Problems/124 - Cube Differences/Solution

by : $124$

 * Cube Differences

Solution

 * $642^3 - 641^3 = 1 \, 234 \, 567$

Proof
We need to find $a$ and $b$ such that $a^3 - b^3 = 1 \, 234 \, 567$.

The assumption is that $a, b \in \Z_{>0}$ so we are not concerned about negative numbers.

First we use Difference of Two Cubes to obtain:
 * $a^3 - b^3 = \paren {a - b} \paren {a^2 + a b + b^2}$

We have that:
 * $1 \, 234 \, 567 = 127 \times 9721$

from which it follows either that:


 * $a - b = 127$ and $a^2 + b^2 + a b = 9721$

or:
 * $a - b = 1$ and $a^2 + b^2 + a b = 1 \, 234 \, 567$

Trying the first of these:

However, the above quadratic does not have integer solutions.

So we try:

It is the positive solution we need here, which leads to:


 * $a = 642$
 * $b = 641$