I read in "Letters to a young mathematician" that 4900 is the only square integer that is the sum of consecutive squares (I believe he meant by that "starting from 1", but maybe that's not even necessary). I did a quick run through with a python script and of course this seems totally devoid of a computational pattern. Why is 4900 (and 1 of course) the only number such that this works?


I did find out that the sum of squares is the following...


$\sum^{n}\_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}$



4900, a particularly square number
This is a classical Diophantine equation (Mordell, *Diophantine Equations*, p. 258). Apart from n = 0, 1, -1, there is only the solution n = 24. Proofs by G. N. Watson (1919), W. Ljunggren (1952).


There is some history in <http://www.math.ubc.ca/~bennett/paper21.pdf> .


