Lucas: Q. 1180

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 $x = 24$.