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

From ProofWiki
Jump to navigation Jump to search

Puzzles and Curious Problems by Henry Ernest Dudeney: $124$

Cube Differences
If we wanted to find a way of making the number $1 \, 234 \, 567$ the difference between two squares,
we could of course write down $517 \, 284$ and $617 \, 283$ --
a half of the number plus $\tfrac 1 2$ and minus $\tfrac 1 2$ respectively to be squared.
But it will be found a little more difficult to discover two cubes the difference of which is $1 \, 234 \, 567$.


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:

\(\ds a - b\) \(=\) \(\ds 127\)
\(\ds a^2 + b^2 + a b\) \(=\) \(\ds 9721\)
\(\ds \leadsto \ \ \) \(\ds a^2 + \paren {a - 127}^2 + a \paren {a - 127}\) \(=\) \(\ds 9721\) substituting $b = a - 127$
\(\ds \leadsto \ \ \) \(\ds 3 a^2 - 3 \times 127 a + 127^2\) \(=\) \(\ds 9721\)
\(\ds \leadsto \ \ \) \(\ds 3 a^2 - 381 a + 6408\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds a^2 - 127 a + 2136\) \(=\) \(\ds 0\) simplifying

However, the above quadratic does not have integer solutions.


So we try:

\(\ds a - b\) \(=\) \(\ds 1\)
\(\ds a^2 + b^2 + a b\) \(=\) \(\ds 1234567\)
\(\ds \leadsto \ \ \) \(\ds a^2 + \paren {a - 1}^2 + a \paren {a - 1}\) \(=\) \(\ds 1234567\) substituting $b = a - 1$
\(\ds \leadsto \ \ \) \(\ds 3 a^2 - 3 a + 1\) \(=\) \(\ds 1234567\)
\(\ds \leadsto \ \ \) \(\ds 3 a^2 - 3 a - 1234566\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds a^2 - a - 411522\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {a + 641} \paren {a - 642}\) \(=\) \(\ds 0\) factoring
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds -641 \text { or } 642\) factoring

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

$a = 642$
$b = 641$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Puzzles_and_Curious_Problems/124_-_Cube_Differences/Solution&oldid=561734"