# 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.