When does a square equal to the sum of k consecutive squares?

Sum of consecutive squares = ( ( n * ( n + 1 ) * ( 2n + 1 ) ) / 6)

Lucas claimed that the only solutions were k = 1 and k = 24

If k = 1 => 1^2 = 1^2

If k = 24 => 1^2 + 2^2 + 3^2 + ... + 22^2 + 23^2 + 24^2 = 4900 = 70^2

A more general problem is to determine the set S of values k for which there exists a square equal to the sum of k consecutive squares.

For example,

25^2 + 26^2 + 27^2 + ... + 623^2 + 624^2 = 9010^2

More research is needed.

http://www.jstor.org/pss/2974904