I asked this question at math.SE a couple of months ago and only got a partial answer, so I thought I would try here.

It is known that, for $n \geq 5$, it is possible to partition the integers $\{1, 2, \ldots, n\}$ into two disjoint subsets such that the product of the elements in one set equals the sum of the elements in the other. One solution is the following:

Let $N = \{1, 2, \ldots, n\}$.

If $n$ is even, take $P = \{1, \frac{n-2}{2}, n\}$ and $N-P$ as the two sets.

If $n$ is odd, take $P = \{1, \frac{n-1}{2}, n-1\}$ and $N-P$ as the two sets.

My question is this:

Is this partition unique for infinitely many $n$?

Background: The problem of proving that the partition is possible was posed several years ago as Problem 2826 in the journal *Crux Mathematicorum*, with solutions in the April 2004 issue. Every one of the 20 or so solvers (including me, which is why I'm interested in the question) came up with the partition given here. The person who originally posed the problem also asked if the partition is unique for infinitely many $n$. I don't think anyone ever submitted an answer to that latter question to *Crux* (although I cannot verify that, as I no longer have a subscription). I thought someone here might be able to give an answer.

The partial answers to the math.SE question were

1) Matthew Conroy showed by brute force calculation that, for $5 \leq n \leq 100$, the only values of $n$ that have only this solution are $5,6,7,8,9,13,18,$ and $34$.

2) Derek Jennings showed that for $n=4m$ we can obtain a partition with the required property by taking $P=\{8,m−1,m+1\\}$ for $m>1$ and $m \neq 7$ or $9$. Thus the partition in the question is not unique for $n$ a multiple of $4$ and greater than $36$.