Sum Equal Product


There are many articles in re this issue on the Web, although many are restricted to special cases (e.g., John Cook’s sum of tangents = product of tangents), and even some involving mixed numbers. 


There is one excellent article in re: sum equal product written by two Polish mathematicians which is replete with wonderful proofs that I am not sophisticated enough to understand. (Kurlandchick & Nowicki, Nov `98)


Actually, my question is quite straightforward. Is there a formula or some kind of algorithm that can show which sums of positive, unequal integers equal their products?


Falco



Sum Equals Product
Let the positive, unequal integers be $a\_1 < a\_2 < \cdots < a\_k$ with $a\_1+a\_2+\cdots+a\_k = a\_1a\_2\cdots a\_k = n$. Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$. Note that $a\_k \ge k$. Then we have $a\_1 + a\_2 + \cdots + a\_k < ka\_k$, while $a\_1a\_2\cdots a\_k \ge (k-1)!a\_k$. So we must have $ka\_k > (k-1)!a\_k$, which means that $k > (k-1)!$. This means $k \le 3$. If $k = 2$, then we have $a\_1 + a\_2 = a\_1a\_2$, or $\frac{1}{a\_1}+\frac{1}{a\_2} = 1$. This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a\_i$ be unequal.


The case $k = 3$ is then the only tricky case. If $a\_1 > 1$, then we have $a\_1+a\_2+a\_3 < 3a\_3$ and $a\_1a\_2a\_3 \ge 6a\_3$. So $a\_1 = 1$. We then must find solutions to $1 + a\_2 + a\_3 = a\_2a\_3$. This can be rewritten as $(a\_2-1)(a\_3-1) = 2$, which as $a\_2 < a\_3$ are integers immediately gives $a\_2 = 2, a\_3 = 3$.


In summary, all possible solutions are $\{n\}$ for all positive integers $n$ and $\{1,2,3\}$.


History: this is definitely a classic problem; see for example [2006 USA Mathematical Olympiad problem #4](http://amc.maa.org/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2006.pdf%20%222006%20USA%20Mathematical%20Olympiad%20problem%20#4%22) (pdf, problem is on second page), which is your problem with several restrictions removed.


