*This is my translation of Lucas’s 1877 proof that a square pyramid of balls will only have a square number of balls if the base side is twenty-four.*

“Solutions to questions posed in The New Annals: Question 1180.” M. Édouard Lucas, Nouvelles annales de mathématiques. Journal des candidats aux écoles polytechnique et normale, second series, volume 16 (1877), 429-432 (See second series, volume 14, p. 336)

A pile of balls with a square base does not contain a number of balls equal to a square of a whole number unless each side of its base contains twenty-four balls. (Édouard Lucas)

Indeed, we know that the sum of the squares of the first \(x\) whole numbers is equal to the expression \[\frac{x(x+1)(2x+1)}{6}\] From this we can posit \[x(x+1)(2x+1)=6y^2\] but the factors \(x, x+1\) and \(2x+1\) are co-prime, so the preceding equation can be composed the following nine ways:

1 | \(x=6u^2\) | \(x+1=v^2\) | \(2x+1=w^2\) |

2 | \(x=3u^2\) | \(x+1=2v^2\) | \(2x+1=w^2\) |

3 | \(x=3u^2\) | \(x+1=v^2\) | \(2x+1=2w^2\) |

4 | \(x=2u^2\) | \(x+1=3v^2\) | \(2x+1=w^2\) |

5 | \(x=2u^2\) | \(x+1=v^2\) | \(2x+1=3w^2\) |

6 | \(x=u^2\) | \(x+1=6v^2\) | \(2x+1=w^2\) |

7 | \(x=u^2\) | \(x+1=3v^2\) | \(2x+1=2w^2\) |

8 | \(x=u^2\) | \(x+1=2v^2\) | \(2x+1=3w^2\) |

9 | \(x=u^2\) | \(x+1=v^2\) | \(2x+1=w^2\) |

**Case 1:** We have \[w^2 = 12u^2\] and, since the factors \(w+1\) and \(w-1\) have a greatest common denominator of two, we can conclude, allowing for negative values of \(w\), \[w+1 = 2a^2\] but, furthermore, \[w^2+1 = 2v^2\]

The last two equations can be solved at the same time. This system of equations was treated completely by M. Gerono (in this volume, p. 231); it only allows for complete solutions when \(w=\pm 1\) and \(w=\pm 7\). These values are both valid in the first equation, so we can deduce that \(x = 0\) or \(x = 24\). Thus \[1^2 + 2^2 + 3^2 + … + 24^2 = \frac{24 \cdot 25 \cdot 49}{6} = 4900\]

**Case 2:** This hypothesis leads to the equation \[2v^2 – 3u^2 = 1\] which is impossible mod 3.

**Case 3:** From this composition we can derive the equation \[2w^2 – 6u^2 = 1\] which is impossible mod 2.

**Case 4:** We easily derive \[w^2 + 1 = 6v^2\] an impossible equation mod 3.

**Case 5:** This hypothesis gives us the equation \[4u^2 + 1 = 3w^2\] which is impossible both mod 3 and mod 4.

**Case 6:** We find the equation \[6v^2 = u^2 + 1\] which is impossible mod 3.

**Case 7:** We find the equally impossible \[3v^2 = u^2 + 1\]

**Case 8:** This hypothesis only gives the solution x = 1, as follows from a comment at the end of the preceding article.

**Case 9:** This leads to the impossible \[2u^2 + 1 = 6w^2\]

Hence, in conclusion, the sum of the squares of the first \(x\) numbers is never equal to a perfect square, except when \(x = 24\).

Pingback: MEYL: Q. 1194 | Curious Cheetah