MEYL: Q. 1194

This is my translation of Meyl’s 1878 proof that a triangular pyramid of balls will only have a square number of balls if the base side is two or forty-eight.

“Solutions to questions posed in The New Annals: Question 1194.” A. J. J. Meyl, former artillary captain at the Hague, Nouvelles annales de mathématiques. Journal des candidats aux écoles polytechnique et normale, second series, volume 17 (1878), 464-467 (See second series, volume 15, p. 144)

A pile of balls with a triangular base only contains a number of balls equal to the square of a whole number if it has a base side of 1, 2, or 48.

Given \(n\) as the number of balls in the side of a base, the resulting equation is \(\frac{n(n+1)(n+2)}{6} = Q^2\). We assume:

First case: That \(n\) is an even number \(2m\). The equation becomes \(\frac{2m(2m+1)(2m+2)}{6}=Q^2\) and, in removing the square factor 4 from the first member, \[\frac{m(2m+1)(m+1)}{6}=Q^2\]

Now, in the solution to the question 1180 (second series, volume 16, p. 429), M. E. Lucas demonstrated that the only whole number solutions for \(m\) in the last equation are \(m=1, m=24\). Consequently, we have \(n=2, n=48\).

Second case: That \(n\) is an odd number. There are three situations to consider, depending on whether \(n, n+1,\) or \(n+2\) is divisible by 3. For the sake of simplicity, we replace \(n\) in each situation respectively by \(6m+3, 6m-1,\) and \(6m+1\).

Let \(n\) be divisible by 3, such that \(n=6m+3\). The resulting equation is: \[(6m+3)(6m+4)(6m+5) = 6Q^2\] that is, \[(2m+1)(3m+2)(6m+5)=Q^2\]

The factors of the left hand side are co-prime, meaning \[2m+1=x^2, 3m+2=y^2, 6m+5=z^2\] The first two equations give us \[3x^2+1=2y^2\] which is an impossible equation because \[y \equiv (-1,0,+1) \text{(mod 3)}\]

If \(n+1\) is divisible by 3, we have \(n=6m-1\). The equation becomes \[(6m-1)6m(6m+1)=6Q^2\] and, like above, \[6m-1=x^2, 6m=6y^2, 6m+1=z^2\] with the first two giving \[x^2+1=6y^2\] which is impossible because \[x \equiv (-1,0,+1) \text{(mod 3)}\]

Let \(n+2\) be divisible by 3, or \(n = 6m+1\); we have the equation \[(6m+1)(6m+2)(6m+3) = 6Q^2\] and consequently \[6m+1=x^2, 6m+2=2y^2, 6m+3=3z^2\] hence \[(1) \, x^2+1 = 2y^2, 2y^2+1=3z^2, x^2+2=3z^2\] from which it follows \[(2) \, 3z^2=4y^2=(2y+x)(2y-x)\]

Since \(x, y,\) and \(z\) are co-prime, the factors \(2y+x\) and \(2y-x\), due to their sum of \(4y\) and their difference of \(2x\), can only have a common factor of 2. But, \(x\) being odd, this factor does not exist; thus, \(2y+x\) and \(2y-x\) are co-prime. We can posit \[z=pq, 2y+x=3p^2, 2y-x=p^2\]

The values of \(x, y,\) and \(z\) obtained by these equations and substituted into one of the equations in (1) gives us the equation \[9p^4-18p^2q^2+q^4+8=0\] Solving for \(p^2\) gives us \[p^2=q^2 \pm \frac{2}{3}\sqrt{2(q^4-1)}=q^2\pm\frac{2}{3}\sqrt{2(q^2+1)(q+1)(q-1)}\]

The factors in the radicand must have a common factor of two, since \(q\), being a factor of the odd number \(z\), is also odd. We can therefore put this radical in the form \[\sqrt{16\times\frac{q^2+1}{2}\times\frac{q^2-1}{4}}\] which has the co-prime factors \(\frac{q^2+1}{2}\) and \(\frac{q^2-1}{4}\).

In order for \(p^2\) to be rational, the radicand must be a perfect square or cancel itself out; the first case is impossible, because \(q^2-1\) is never square. The second case is only possible when \(q^2 = 1\). In this case, we have successively \(p^2=q^2=1, x=y=a=1, m=0,\) and finally \(n=1\). Thus we can only find the values 1, 2, and 48 for \(n\). This is what was to be proven.

Editor’s note: The factors \(2y+x, 2y-x\) of the right hand side of the equation (2) \[3z^2=(2y+x)(2y-x)\] are co-prime, giving \[z=pq, 2y+x=3p^2, 2y-x=p^2\] or \[z=pq, 2y+x=p^2, 2y-x=3p^2\] but the second hypothesis leads to the same result as the first. The same question was resolved by M. Moret-Blanc.

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