2

OEIS A000330 defines a square pyramid number $a(n)$ as one that evaluates to $$\sum_{a=0}^n{a^2}=\frac{n(n+1)(2n+1)}6$$ The first several solutions are $${0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370,...}$$

As you can see, the only solutions on this list for which $a(n)$ is itself a perfect square are when $n=0,a(n)=0$, $n=1,a(n)=1$, and $n=24,a(n)=4900$. In fact, the first comment on the OEIS page claims that these are the only such numbers in this sequence, a claim later confirmed in 2013 by Watson.

Is there a mathematical proof of this, beyond brute force guess-and-check?

DonielF
  • 1,126
  • 1
  • 8
  • 22
  • https://math.stackexchange.com/questions/74265/the-sum-of-the-first-n-squares-is-a-square-a-system-of-two-pell-type-equation – zwim Oct 17 '17 at 22:46
  • https://en.wikipedia.org/wiki/%C3%89douard_Lucas and the "cannonball" problem. The easiest proof is that by Anglin, listed in the Lucas page. However, even that one does not strike me as entirely elementary. – Will Jagy Oct 17 '17 at 23:34
  • https://en.wikipedia.org/wiki/Cannonball_problem – Will Jagy Oct 17 '17 at 23:35

0 Answers0