The following is a puzzle of Dudeney:
Professor Rackbrane pointed out one morning that the cubes of successive numbers, starting from $1,$ would sum to a square number... He stated that if you are forbidden to use the $1,$ the lowest answer is the cubes of $23,24,25,$ which together equal $204^2.$ He proposed to seek the next lowest number using more than three consecutibe cubes and as many more as you like excluding $1.$
There is an answer provided but a proof is not included:
The cubes of $14,15,$ up to $25$ inclusive (twelve in all) add up to... the square of $312.$ The next lowest answer is the five cubes of $25,26,27,28,$ and $29,$ which together equal $315^2.$
My query is more general: Classify all finite sets of consecutive positive integers, the sum of whose cubes is a square. Any idea how we can do this? If no one manages to answer this question, I will accept an answer that shows how Dudeney came to the minimal solutions.
Here are my thoughts on the matter so far: Clearly the sum of first $n$ positive cubes works. The sum of the cubes of $m+1,m+2,\ldots,n$ for positive $n$ and non-negative $m$ is $$(1^3+2^3+\cdots+n^3)-(1^3+2^3+\cdots+m^3) = \left[\frac{n(n+1)}{2}\right]^2-\left[\frac{m(m+1)}{2}\right]^2.$$ If this is equal to a square, it is equivalent to seeking all Pythagorean triples such that at least two of the elements of the triple are triangular numbers. At that point, I tried to use the classification of all (primitive) Pythagorean triples, but that went nowhere.