# 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 $$x = 24$$.

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