This is a shorter post, about a silly little problem I came up with a few months ago. It’s not a very intelligent problem, in that it doesn’t really serve any mathematical purpose; nor is it altogether tough; but I think it has a certain aesthetic pleasantness to it.

Problem. Let $\def\s#1{(\mathrm S_{#1})}\s{15}$ be the following system of 15 linear Diophantine equations in 15 variables.

Does this system have an integer solution?

More generally, if we define $\s n$ analogously for $n \ge 1$, what is the greatest $n$ for which there exists an integer solution?

If you want to try it for yourself, don’t read the next bit because I’ll also say some words about how to solve it. I think it’s kinda fun to discover the curious properties of $\s n$, so easy as it is to just keep reading, consider giving it a shot yourself.

So how would you approach such a problem? Well, writing it out in full would help. Then if you look closely, you’ll notice some interesting relations. Here’s a useful one:

This implies that $I = E$. Thus, $9 = N + I + N + E = 2(N + I)$ for any integral solution, which is impossible. So $\s{15}$ is has no integral solutions.

In fact, this helps us for the second part, because we have shown that even $\s{13}$ has no integral solutions. Let’s how far down we can go. First, note that $12 = T + W + E + L + V + E$ is redundant, because of the following anagrammatical fact:

So we can do $\s{12}$ iff we can do $\s{11}$.

Well, we can get a bit sneakier with our reductions. $11 = E + L + E + V + E + N$ is the only equation to use the variable $L$, so if we can solve the rest, then we can take $L = 11 - 3E - V - N$. Following this reduction strategy, we can do it again on $(2,W)$, $(4,U)$, $(6,X)$, and $(8,G)$ to leave the equations $\{1,3,5,7,9,10\}$.

But now we can reduce again, and we can consider variables that were occupied by already-eliminated equations! Since we have eliminated equations $\{2,4,6\}$, we can eliminate $(1,O)$, $(3,R)$, $(5,F)$, and $(7,S)$, which pares the problem down to the two equations $9 = N + I + N + E$ and $10 = T + E + N$.

That’s easy to solve, so we conclude that the answer to the problem is $n = 12$.