Henry Ernest Dudeney/Puzzles and Curious Problems/122 - A Question of Cubes/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $122$

A Question of Cubes
From Sum of Sequence of Cubes, the cubes of successive numbers, starting from $1$, sum to a square number.
Thus the cubes of $1$, $2$, $3$ (that is, $1$, $8$, $27$), add to $36$, which is the square of $6$.
If you are forbidden to use the $1$, the lowest answer is the cubes of $23$, $24$ and $25$, which together equal $204^2$.
What is the next lowest number, using more than three consecutive cubes and as many more as you like, but excluding $1$?


Solution

$\ds \sum_{n \mathop = 14}^{25} n^3 = 14^3 + 15^3 + 16^3 + \cdots + 25^3 = 97 \, 344 = 312^2$


The next one after that is:

$\ds \sum_{n \mathop = 25}^{29} n^3 = 25^3 + 26^3 + 27^3 + 28^3 + 29^3 = 99 \, 225 = 315^2$


Proof


This theorem requires a proof.
In particular: I'm surprised we haven't got this in already. I would have expected David Wells to be all over this, but it seems he missed it. Unless he did, we do have it, and I'm missing it now.
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Sources

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