The Mysterious Tetrahedral Squares - An Adventure in Number Theory

Hey! Choo Ray here.

I've been involved in giving sessions for the SIMO National Team and other trainings recently, so naturally I have to hunt for suitable problems. I feel that I've been looking at more problems recently than I have in the lead-up to my participations in IMO/other competitions! Of course, I spend less time on each problem, as my objectives are geared towards discovering problems with good ideas and instructive value rather than solving them myself. However, sometimes I find myself being led down a long rabbit-hole of theory that I apparently ought to know about. In this post I'd like to share about one of these experiences.

One day, I was browsing contest collections on Art Of Problem Solving (AoPS). A question from the 2020 Bulgarian National Olympiad caught my eye:

P4. Are there positive integers $m>4$ and $n$, such that
a) ${m \choose 3}=n^2$
b) ${m \choose 4}=n^2+9$

I clicked on the link, thinking that it seems a rather routine problem. Little did I know what I was getting myself into...

An Unexpected Side Path

At first, everything seemed normal for a supposedly "easy" problem 4.

For part (a), one simply needed to come up with a construction. A clever modulo consideration, coupled with some algebraic manipulation or casework, quickly disposed of part (b).

The question required only the most basic techniques, applied in a prudent manner. I was ready to move on, but a tiny detail nagged at me.

You see, a typical Diophantine problem with simple statement tends to have just a few small solutions. Usually the task is to prove that these small solutions are the only ones.

But part (a) only required a construction, not a proof. Not only that, but the construction that AoPS user VicKmath7 provided for part (a) was $m=50, n=140$. This was surprisingly large to me. Surely if there was a smaller construction, someone else would point it out.

My suspicion that no smaller solution could be found was quickly confirmed by user Solar Plexsus, who wrote

$(1) \;\; {m \choose 3} = n^2$
is wellknown in the theory of figurate numbers. The solutions of equation (1) give us the numbers which are both square and tetrahedral (see MathWorld entry).
The only solutions of equation (1) are $(m,n) = (3,1), \: (4,2), \: (50,140)$ (proven by Meyl in 1878).

Alright, wellknown. It is usually advised not to take such claims by unknown strangers on the internet seriously. A Diophantine having just three solutions, with $(50,140)$ being one of them? Extraordinary, almost too good to believe. Yet everything about this post rang true, with the official-sounding hyperlink and specific reference to the first known solve. Eagerly, I clicked on the link to find out more about these mysterious "tetrahedral numbers".

Building Things Up

The term "tetrahedral number" refers to how such numbers arise from counting spheres stacked in a tetrahedral pattern.

It is rather straightforward to show why these correspond to the binomials ${m \choose 3}$. Simply computing algebraically, the $n$th layer has $$1+2+...+n=\frac{n(n+1)}{2}$$ spheres so the number of spheres in a tetrahedron with $n$ layers should be 

$$ \sum_{k=1}^{n}\frac{k(k+1)}{2} = \sum_{k=1}^{n}\frac{k^2}{2} + \sum_{k=1}^{n}\frac{k}{2} =\frac{n(n+1)(2n+1)}{12} +\frac{n(n+1)}{4} = $$

$$ \frac{n(n+1)}{4}(\frac{2n+1}{3}+1) = \frac{n(n+1)}{4}(\frac{2n+4}{3}) = \frac{n(n+1)(n+2)}{6} $$

Alternatively, one could employ a combinatorial argument like such: Suppose we are picking $3$ numbers from ${1,2,...,n+2}$. If the largest is $k+2$ ($1\le k\le n$) we look at the $k$th layer of the tetrahedron. Then, if the second largest is $m+1$ ($1\le m\le k$) we select the $m$th row of that layer, with $m$ spheres. These correspond exactly to the $m$ choices left for the smallest number.

Let us return to the problem of finding tetrahedral numbers that are also square. Multiple webpages pointed out the that the only tetrahedral numbers which are also square are the $1$st, $2$nd and $48$th. They credited a certain A.J.Meyl for being the first to conquer this problem, just like Solar Plexsus had. It was slightly more difficult to look for Meyl's proof, but eventually, in a Math Stackexchange post, I found a link to exactly what I was looking for: the original proof by Meyl himself.

Excusez mon français

Upon opening the paper I was quickly confronted with a new problem. The talented Meyl had evidently failed to predict the predominant working language of our time. He had made the unfortunate decision to publish his proof in French!

I tried my best for a few agonising minutes to follow his argument about arrangement of "boulets" with "base triangulaire", until suddenly the realisation hit me that $\exists$ translation software . This simplified the process considerably and I could roughly follow the argument:

(Adapted English Translation of Meyl's Proof)

1. To solve $n(n+1)(n+2)=6Q^2$, we split into cases based on the parity of $n$. When $n$ is odd, $n,n+1,n+2$ are pairwise coprime. So each prime power dividing $6Q^2$, which will be $2^{2d}$, $3^{2e}$ or $p^{2f+1}, p>3$, must exactly divide one of the three factors. It follows that the squarefree part of each factor is $1,2,3$ or $6$.

2. If $n\equiv 3\pmod{6}$ then the squarefree parts must distribute as $$n=3x^2, n+1=2y^2, n+2=z^2$$ but then $$z^2\equiv 3x^2+2\equiv 2\pmod{3}$$ which is impossible. If $n\equiv 5\pmod{6}$ then $$n=x^2, n+1=6y^2, n+2=z^2$$ but then $$x^2\equiv 6y^2-1\equiv 2\pmod{3}$$ is again impossible.

3. If $n\equiv 1\pmod{6}$ we have $$n=x^2, n+1=2y^2, n+2=3z^2$$ with $x,y,z$ coprime. We would like to get information from the rigid properties of squares by eliminating $n$, so we establish $$2n+2=4y^2=x^2+3z^2$$ This loses a little information, since we have a relation between all of $x,y,z$ when we had relations between any two. On the flip side, we get a good structure to work with. We realise that we can create the difference of two squares $$4y^2-x^2=(2y-x)(2y+x)=3z^2$$ Any prime dividing both $2y-x$ and $2y+x$ must divide their sum $4y$ and difference $2x$, but $x$ and $y$ share no common factors so it must be $2$. However, this is also not possible as $n=x^2$ was odd so $2y-x$ and $2y+x$ are both odd. Thus $\gcd(2y-x,2y+x)=1$.

4. Why we bothered checking that $2y-x, 2y+x$ are coprime is because it allows us to apply again the argument about squarefree parts when we look at the equation $(2y-x)(2y+x)=3z^2$. Here $2y-x$ and $2y+x$ can only "inherit" from $3z^2$ prime powers of the form $p^{2k}, p\neq 3$ , or $3^{2k+1}$. Thus their squarefree parts must be 1 and 3.

5. Are we ready for more variables? (How else are we going to use this new information?) Put for example $$2y-x=a^2, 2y+x=3b^2$$ so that $z=ab$. Now it seems that we have run out of sand for our sandcastle until we remember that earlier we saved ourselves a bucket of information when eliminating $n$ and forming an equation with $x,y,z$. We can consider the relation between just $y,z$ for further progress.

6. With $3z^2=n+2=2y^2+1$, we can substitute $$3a^2b^2=3z^2=2y^2+1=2(\frac{a^2+3b^2}{4})^2+1$$ where we used $$4y=(2y-x)+(2y+x)=a^2+3b^2$$ Note that we would get the same equation even if the squarefree parts were switched (that is, $2y-x=3b^2, 2y+x=a^2$).

7. Now we may put on the finishing touches. Simplifying, $$3a^2b^2=2(\frac{a^2+3b^2}{4})^2+1 $$ $$24a^2b^2=(a^2+3b^2)^2+8 $$ $$ 24a^2b^2=a^4+6a^2b^2+9b^4+8 $$ $$ a^4-18a^2b^2+9b^4+8=0 $$ $$ 9a^4-18a^2b^2+9b^4=8a^4-8 $$ $$ (a^2-b^2)^2=\frac{8}{9}(a^4-1) $$ $$ b^2-a^2=\pm \frac{2}{3}\sqrt{2(a^4-1)} $$ Here I used completing the square for my manipulation, but all this is really just the standard derivation of the quadratic formula if we observe that we have a quadratic in $a^2, b^2$. Anyway, with $2(a^4-1)=2(a^2+1)(a^2-1)$ being both an integer and a square of a rational, it must be the square of an integer. Now considering $a^4-1$ is even and $a$ is clearly odd. Then $a^2+1\equiv 2\pmod{4}$, $\gcd(2(a^2+1),a^2-1)=4$ and we can apply our argument on squarefree parts once again, this time to tell us that both $2(a^2+1)$ and $a^2-1$ are both squares! Trivially the only square that is one less than a square must be zero, giving us $a^2=1$. Tracing backwards, $b^2=1$, then $z=ab=1$, then $n=3z^2-2=1$.

The Secret Around the Corner

Did you get lulled into a false sense of security? Or were you constantly aware in the previous section that we have yet to address the case where $n$ is even? You should have been, since we are supposed to get 3 solutions to $n$ and so far we have gotten just one. Here is how Meyl dealt with the even case.

Let $n=2m$, then $2m(2m+1)(2m+2)=6Q^2$ or $\frac{m(m+1)(2m+1)}{6}=(\frac{Q}{2})^2$.

Or, dans la solution de la question 1180 ($2^e$ série, t. XVI, p. 429), M. E. Lucas a démontré que les seules valeurs en nombres entiers pour m, dans cette dernière équation, sont m = 1, m = 24. Par conséquent, on a n = 2, n = 48.

Of course, the cunning Meyl quoted a previous solution in his proof! I shook my head in disbelief. I tried to see if Meyl's ideas extended to the even case, but ran into some difficulties. Naturally I had to investigate for any surviving record of M. E. Lucas' work. This time I was not so lucky to find the original.

However, in my search I made another fascinating discovery. The observant reader may have noted $\frac{m(m+1)(2m+1)}{6}$ as the formula for the sum of first $m$ squares $1^2+2^2+3^2+...+m^2$. Then, we may interpret this geometrically. Just as we did with tetrahedral numbers, where we built triangular layers of spheres, in this case we can build a square pyramid instead! Following a series of clues led me to a 2018 paper by W.S.Anglin titled "The Square Pyramid Puzzle". In it I read the following hilarious but upsetting line:

In 1877, having pointed out that there was a gap in Moret-Blane's argument, Lucas published a proof which also contained a gap.

As it turns out, our hero Meyl had inadvertently trusted an incomplete proof of a fellow Frenchman. Perhaps we may better understand this oversight if we took into account the fact that, in another twist, M. E. Lucas is Monsieur Édouard Lucas, a prominent French mathematician of the time (of "Lucas numbers" fame).

In Anglin's paper, which claimed to use only elementary techniques with a proof "so short and simple that it is surprising that it escaped mathematicians for over a hundred years", it is written that only in 1918 had a complete proof of the square pyramid problem been given using elliptic functions.

Certainly Anglin's methods (infinite descent, Pell's equation theory, quadratic residue theory) are more elementary when compared to elliptic functions, but to me, the proof is by no means simple to follow. I spent about three days to digest the paper.

Closing Thoughts

If A.J.Meyl accidentally used an unproven result in his solution, does that mean that he did not produce a valid proof himself? Was AoPS user Solar Plexsus wrong all along, together with the many webpages attributing the first solution to Meyl?

This tale highlights the importance of proofreading and peer review in mathematics. Like the pyramid structure, mathematical proofs often build on layers upon layers of previous work. From time to time, we need to inspect the middle for defects, or we risk a nasty surprise should everything come crashing down.

Part of it may have to do with how we record our proofs. Sometimes we choose to omit certain details for convenience, or perhaps we think that having a series of long calculations may turn the reader away. But from my experience, there is nothing more discouraging than having an important lemma's proof "left to the reader". Worst still, such remarks often come with no clues about how one might go about proving it. These omissions severely hamper the work of anyone trying to build upon existing results. I believe that in our mathematical writing, we should provide a layer of explanations beyond what our target audience is expected to know. We should keep in mind that whatever results we take for granted now, we also viewed as shrouded in mystery once.

I hope you enjoyed reading about how I stumbled upon a rich slice of mathematical history. Let me know if you liked it and if you have any suggestions on how I may improve / what you want to see for my next post. I am quite new to blogging so any advice would be helpful. Thanks!

A.J.Meyl's original proof in French