The Sum of Perfect Squares

Question

In Symmetry and the Monster, I ran across this interesting fact:

Let $\displaystyle f(n) = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \sum_{k=1}^{n} k^2$

Let $x$ be an integer

Then $f(n) = x^2$ for only two tuples $(n,x)$: $(1,1)$ and $(24,70)$

How would you prove this? Intuitively, is there something special about $24$?

The identity $f(24) = 70^2$ can be explained with the Leech lattice but it isn't clear to me how this helps proving that there is no other such identity. — t.b., May 03 '11 at 15:53
It can also be found using elliptic curves, also there is an elementary proof but it is two pages of hard work. — quanta, May 03 '11 at 15:58
See also: The sum of the first $n$ squares is a square: a system of two Pell-type-equations. — Martin Sleziak, Apr 13 '19 at 23:13

score 8 · Accepted Answer · answered May 03 '11 at 15:53

8

Here is an article that proves this fact (hopefully it is accessible). Also, there just usually aren't many numbers that are more than one kind of figurate number at once, and 24 happens to be the only solution when we look for square pyramidal numbers that are also square numbers.

answered May 03 '11 at 15:53

Zev Chonoles

1

Here is JSTOR link to the article: http://www.jstor.org/stable/2323911 – Martin Sleziak Jan 20 '16 at 08:22

1 Answers1