Numbers whose Cube equals Sum of Sequence of that many Squares/Mistake

Source Work

 * The Dictionary
 * $103,823$
 * $103,823$

Mistake

 * $103,823 = 47^3 = 22^2 + 23^2 + \ldots + 67^2 + 68^2$, the smallest representation of a cube as the sum of consecutive squares. The next smallest is $2161^3$.

What was omitted from here is the fact that the number of squares is $47$.

That is, this is the smallest $m$ such that $m^3$ equals the sum of $m$ consecutive squares, not just any number of them.

Whether there are indeed any other cubes less than $47^3$ or between $47^3$ and $2161^3$ which are the sum of any number of consecutive squares (apart from the trivial $6$th powers) needs to be established.