(related to Problem: The Stonemason's Problem)
The puzzle amounts to this. Find the smallest square number that may be expressed as the sum of more than three consecutive cubes, the cube $1$ being barred. As more than three heaps were to be supplied, this condition shuts out the otherwise smallest answer, $23^3 + 24^3 + 25^3 = 204^2.$ But it admits the answer, $25^3 + 26^3 + 27^3 + 28^3+ 29^3 = 315^2.$ The correct answer, however, requires more heaps, but a smaller aggregate number of blocks. Here it is: $143 + 153 + \ldots$ up to $253$ inclusive, or twelve heaps in all, which, added together, make $97,344$ blocks of stone that may be laid out to form a square $312 \times 312.$ I will just remark that one key to the solution lies in what are called triangular numbers. (See also Digital Puzzles, A Study in Thrift, and The Stonemason's Problem)
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