Sum equal to the product I know that $2 + 2 =  2 \cdot 2$ and $1 + 2 + 3 = 1 \cdot 2 \cdot 3.$
My question: Are there other positive integers with sum equal to the product? (The number 1 can not appear more than once among them.)
I am therefore excluding the infinity of solutions obtained by repeating 1, such as:
$$1 + 1 + 2 + 4 = 1 \cdot 1 \cdot 2 \cdot 4$$
$$1 + 1 + 1 + 2 + 5 = 1 \cdot 1 \cdot 1 \cdot 2 \cdot 5$$
 A: First we shall look at 2 numbers.
Consider $a,b \in \mathbb{N}$.  We have $ab=a+b$ $$\implies ab-a=b$$ $$\implies a(b-1)=b$$
$$\therefore b-1|b$$ $$\implies b=2$$ $$\implies a=2$$ And this is the only solution for two numbers.
Now let us look at $n$ numbers, $a_1,a_2,a_3, \dots a_n \in \mathbb{N}$. WlOG $a_1 \leq a_2 \leq \cdots \leq a_n$
We have $a_1a_2a_3 \cdots a_{n-1}a_n=a_1+a_2+a_3+ \cdots + a_n$. Clearly the way to optimize the situation (i.e minimizing the L.H.S while maximizing the R.H.S) is to put $a_1=1$ and $a_2,a_3, \cdots a_n=2$
Then $2^{n-1}=2n-1$ $$\implies2^{n-2}<n-1$$ $$\implies n\leq3$$
So now we look at $n=3$ (we have already done $n=2$).
If none of the terms is equal to one the most likely case is 2,2,2 which doesn't work so none can work. So $a_1=1$ then we have $$a_2a_3=1+a_2+a_3$$ $$\implies a_2a_3-a_2=a_3+1$$ $$\implies a_2(a_3-1)=a_3+1$$ $$\implies a_3-1 | a_3+1$$ $$\implies a_3=2,3$$ Which you can sub in and you find the only solution to be $a_1=1,a_2=2,a_3=3$.
