Show that for all odd positive integer $n$, there exists a set $A$ where $A= [a_1, a_2, a_3, ... , a_n]$ and $\displaystyle\sum_{i=1}^n a_i =\prod_{i=1}^n a_i$.
Edit: $a_1,...,a_n$ must be distinct.
Here is my approach:
We can prove this by induction. The base case of $n = 1$ holds since $a_1 = a_1$. Now assume that there exists a set of $a_1,a_2,...,a_{2k-1}$ such that $a_1+a_2 +\cdots+a_{2k-1} = a_1a_2 \cdots a_{2k-1}$ is true for some $k$. If we add $a_{2k}+a_{2k+1}$ to both sides, we then must have numbers $a_2k$ and $a_{2k+1}$ such that $a_1a_2 \cdots a_{2k-1}+a_{2k}+a_{2k+1} = a_1a_2 \cdots a_{2k-1}a_{2k}a_{2k+1}$. We can think of this more simply as $x+y+z = xyz$ where $x = a_1a_2 \cdots a_{2k-1}, y = a_{2k}, z = a_{2k+1}$ and $y \neq z$ . We have that $z = \dfrac{x+y}{xy-1}$. We can keep increasing the $y$ value so that $y$ and $z$ are not equal to any of $a_1,a_2,...a_{2k-1}$ and $y \neq z$. Therefore we have finished the induction.
I am thinking my reasoning in the last step isn't very rigorous. Should I make it more rigorous?