Learn One Weird Trick And Easily Solve The Product-Sum Problem

A tribute to Martin Gardner.

For which sets of positive integers does the sum equal the product? For example, when does $x_1 + x_2 = x_1\cdot x_2$? It's easy to see that this is only true when $x_1 = x_2 = 2$.

In the general case our equality is $\sum_{i=1}^{n} x_i= \prod_{i=1}^{n} x_i$. We can rearrange terms to give $$x_1+x_2+\sum_{i=3}^{n} x_i= x_1\cdot x_2\cdot \prod_{i=3}^{n} x_i,$$ and this in turn factors nicely to give us $$\left( x_1\cdot \prod_{i=3}^{n} x_i - 1\right)\cdot \left( x_2\cdot \prod_{i=3}^{n} x_i - 1\right) = \left( \prod_{i=3}^{n} x_i \right)\cdot \left(\sum_{i=3}^{n} x_i \right) + 1.$$ How does this help? Consider the case $n=5$, and without loss of generality assume $x_i \ge x_{i+1}$ for all $i$. If $x_3=x_4=x_5=1$ our factorized equation becomes $$(x_1-1)\cdot(x_2-1)=4,$$ with the obvious solutions $x_1=5, x_2=2; x_1=3, x_2=3$. The only remaining case to consider is $x_3=2$, as any other case forces  $\sum_{i=1}^{n} x_i < \prod_{i=1}^{n} x_i$. For this case our equality becomes  $$\left( 2 x_1 - 1\right)\cdot \left( 2 x_2 - 1\right) = 9.$$ This gives us the remaining solution $x_1 = x_2 = x_3 = 2$ with the other $x_i = 1$.

This is quite efficient. For example, for the case $n=100$ the only equations we need to consider are $$\begin{aligned} (x_1-1)\cdot(x_2-1) &= 99\\                          (2 x_1-1)\cdot(2 x_2-1) &= 199\\                          (3 x_1-1)\cdot(3 x_2-1) &= 301\\                          (4 x_1-1)\cdot(4 x_2-1) &= 405\\                          (4 x_1-1)\cdot(4 x_2-1) &= 401\\                          (6 x_1-1)\cdot(6 x_2-1) &= 607\\                          (9 x_1-1)\cdot(9 x_2-1) &= 919\\                          (8 x_1-1)\cdot(8 x_2-1) &= 809\\                          (12 x_1-1)\cdot(12 x_2-1) &= 1225\\                          (16 x_1-1)\cdot(16 x_2-1) &= 1633. \end{aligned}$$ Now 199, 401, 607, 919, 809 are all prime ruling them out immediately, and 301 and 1633 don't have factors of the right form. The other equations yield the five solutions $(100,2), (34,4), (12,10), (7,4,4), (3,3,2,2,2)$ with the other $x_i = 1$.

For $n = 1000$ you'd need to consider 52 cases, but most of these are eliminated immediately. I get the six solutions $(1000,2), (334,4), (112,10), (38,28), (67,4,4), (16,4,4,4)$.

Have fun!