Product is Sum, Sum is Product

Let $(a_1, a_2, \ldots a_n)$ denote the solution sets for a value of $n$ .

If $n = 1$ , then it is true for all $a_1 = a_1$ .

If $n = 2$ , then $(2, 2)$ satisfies the equation.

If $n = 3$ , then $(1, 2, 3)$ satisfies the equation.

If $n = 4$ , then $(1, 1, 2, 4)$ satisfies the equation.

If $n = 5$ , then $(1, 1, 1, 2, 5)$ satisfies the equation.

As you increase $n$ , you'll notice that there is a pattern such that:

If $n = k$ for some $k$ , then we can just put $(k-2)$ $1$ 's, a $2$ and $k$ in the parentheses: $(1, 1, 1, \ldots 1, 2, k)$ .

Using this into the equation,

$1 + 1 + 1 + \ldots + 1 + 2 + k = 1 \times 1 \times \ldots \times 1 \times 2 \times k$

$k - 2 + 2 + k = 1^{k-2} \times 2 \times k$

$2k = 2k$

Thus, there is no maximum value.

Note that there may be additional sets that can satisfy a given value $n$ , but we just need a set to prove that a solution exists for that value.