Let $R$ be a commutative ring with identity and $R^n$ be the direct sum of $R$. Find all $a_1, a_2, \cdots, a_n \in R$ such that $$a_1 + a_2 + \cdots + a_n = a_1a_2\cdots a_n,$$ or, in other words, if we let $$ K_n = \left\{ (a_1, a_2, \cdots, a_n) \in R^n \, | \, a_1 + a_2 + \cdots + a_n = a_1a_2\cdots a_n \right\},$$ we are interested in determining the algebraic nature of $K_n$.
Example 1: Let $R = \mathbb{Z}$ and consider $\mathbb{Z}^3$; then we have the classic number theory problem of determining $a,b,c$ such that $$a + b + c= abc.$$ $(1,2,3)$ is one such solution to the problem.
Example 2: Let $R = \mathbb{Z}[i]$ and consider $\mathbb{Z}[i]^3$; then this is a twist of the above example. One such solution is $(1-i, 1+i, 2)$.
For the purposes of notation, we will let $(a_i)$ be shorthand for $(a_i)_{i=1}^n$, $\sum_n a_i$ to denote $\sum_{i=1}^n a_i$, and $\prod_n a_i$ denote $\prod_{i=1}^n a_i$. We will also say $(a_i)$ satisfies the product-sum property if $(a_i) \in K_n$.
By observation, we see that $0_{R^n} = (0_R) \in K_n$, so $K_n$ is non-empty. In addition, if $S_n$ denotes the symmetric group on $\{1, \cdots, n\}$ and we define $\cdot: S_n \times K_n \to K_n$ by $\sigma \cdot (a_i) = (a_{\sigma(i)})$, which is defined since it follows by commutativity in $R$, we find $\cdot$ is a group action of $S_n$ on $K_n$. This means if we know $(a_i) \in K_n$, then this informs us that any permutation of the elements in $(a_i)$ will also be a solution to above property. Furthermore, if we let $Z_R = \{\alpha \in R \, | \, \alpha^n = \alpha\}$, where $n$ is the same integer as indicated in $R^n$, we see, under multiplication $\cdot$ in $R$, that $(Z_R, \cdot)$ is a commutative monoid. Thus, if we define $\cdot: Z_R \times K_n \to K_n$ by $\alpha \cdot (a_i) = (\alpha a_i)$, which is defined since $$\sum_n \alpha a_i = \alpha \sum_n a_i = \alpha \prod_n a_i = \alpha^n \prod_n a_i = \prod_n \alpha a_i,$$ we have $\cdot$ is a monoid action of $Z_R$ on $K_n$. This means if we know $(a_i) \in K_n$, i.e. $(a_i)$ is a solution to the above property, then $(\alpha a_i)$ is also a solution.
Let's write $K_n$ as a union of two sets $A$ and $B$, where $A$ is defined to be collection of elements $(a_i)$ such that $a_i = 0_R$ for at least one $i$, and $B$ is defined to be collection of elements $(a_i)$ such that $a_i \neq 0_R$ for any $i$; one way of thinking about is that the "trivial" solutions are packaged into $A$ and the "non-trivial" solutions are packaged into $B$. Then, by definition, $K_n$ is the disjoint union of $A$ and $B$, i.e. $K_n = A \sqcup B$.
Let's consider $A$; notice $A$ is non-empty since $(0) \in A$. Since at least one of $a_i = 0$, $\prod_n a_i = 0$. In addition, since we have a group action of $S_n$ on $K_n$, without loss of generality, we may assume $a_n = 0$. By this construction, we find $A$ is an additive abelian group; as a consequence, we can build an $R$-module structure on $A$ by defining $r \cdot (a_i) = (ra_i)$. Then an immediate consequence is that $A = \mathrm{Span}_R(\{E\})$, where $$E = \left\{(1,0,\cdots,0,-1, 0), (0,1,\cdots,0,-1, 0), \cdots, (0,0,\cdots,1,-1, 0) \right\}$$ with $|E| = n-2$. Since $E$ spans $A$ and the vectors are linearly independent over $R$, this means $A$ is a free $R$-module of finite rank, so that $A = \langle E \rangle$. Thus, $K_n = \langle E \rangle \sqcup B$.
Let's consider $B$; this structure is MUCH trickier to work with compared to $A$. If $\mathrm{char} \, R = 0$, then we at least know $B$ is non-empty; in particular, we have $(\overbrace{1,\cdots,1}^{n-2 \, \textbf{times}},2,n) \in B $. However, if $\mathrm{char} \, R > 0$, then it's not immediately clear if $B$ is non-empty. That said, if we modify $Z_R$ such that we have $Z_R^* = \{\alpha \in R \, \backslash \{0\} \, | \, \alpha^{n} = \alpha\}$, then, continuing from the previous information given, we have $\cdot$ is a monoid action of $Z_R^*$ on $B$, so that we know if $(a_i) \in B$, then so is $(\alpha a_i) \in B$.
I am not sure how to proceed from here. Instead, I made additional assumptions on $R$; if $R$ is a unique factorization domain, then we have a way of classifying the elements in $B$, i.e. $(-1 + \prod_{n-1} a_i)a_n$ can be written as a product of powers of irreducibles with a unit. Since factorization in a UFD is...well, unique, that means we can associate such irreducibles to the elements $(-1 + \prod_{n-1} a_i)$ and $a_n$ and then determine $a_i$ from then on. However, if this assumption is alleviated, it is not clear how to determine $(a_i)$ in $B$. So this begs the following questions.
Question 1: If we don't impose any assumptions on $R$, how can we determine $B$?
Question 2: If $\mathrm{char} \, R > 0$, is $B$ always non-empty?
Update 1 (11/6/2019): As damiano had pointed out, if $R = \mathbb{F}_2$ with $n=2$, then $\mathrm{char} \, R > 0$ and the only points belonging $K_2$ is $(0,0)$, so that $B$ is empty, settling Question 2.
Update 2 (11/7/2019): Vic Camillo of University of Iowa and I were able to prove the following:
Theorem Let $R$ be a commutative unital ring and consider $K_2$. Then the elements $(a,b) \in K_2$ are precisely those that satisfy $a = 1+u$ and $b = 1+u^{-1}$, where $u \in U(R)$ and $U(R)$ are the group of units of $R$.
Proof Since $K_2$ is non-empty, there exists $(a,b) \in K_2$ such that $a+b = ab$. Since $R$ is a commutative ring with identity, the expression can be rewritten as $1 = (a-1)(b-1)$ by realizing we have \begin{align*} 0 &= ab-a-b \\ &= a(b-1) - b \\ &= a(b-1)-b+1-1 \\ &= (a-1)(b-1) -1. \end{align*} This means, by definition of a unit, $a-1 \in U(R)$, so $a = 1+u$ for $u \in U(R)$. Then, by the above equation, we obtain $b = 1+v$, where $v \in U(R)$ such that $uv = 1$, where $u$ is as previously stated. Denoting $v = u^{-1}$, we have $b = 1+u^{-1}$. Thus, the set of elements $(a,b) \in K_2$ are precisely those that satisfy $a = 1+u, b = 1+u^{-1}$, where $u \in U(R)$, as desired.
Thus, the structure for $K_2$ is $K_2 = \langle E \rangle \sqcup B = (0) \, \sqcup G$, where $$G = \{(a,b) \in (R\backslash \{0\})^2 \, | \, a = 1+u, b=1+u^{-1}, u \in U(R)\}.$$
