Contest Math: construct equations with$ \sum_{i=1}^k a_i = \prod_{j=1}^ka_j$ where $a_k$ are all positive rational number. The question is to find all possible integer n such that there exists at least 2 positive rational numbers $a_k$ such that $n =  \sum_{i=1}^k a_i = \prod_{j=1}^ka_j.$
What I think is obvious is, for any composite numbers >=6, they all qualify because as long as you can find a pair of $a_1$ and $a_2$ whose $* = n$ but $+\leq n$, you can use $*1$ to make addition side up.
However, the remaining is quite hard, because $a_n$ can be non-integer, my current guess is as long as you use some 1/n, you need to pay back n*times. But I stuck here for like a day...
 A: I just got something more, now, for any even number >=4, it's obviously all of them can be written as such $2*n*1$...as many as you need...$*1$, for odd number >= 9, they are also fine. This is because, for any n, you can always write it as $n/2*(4)*(1/2)*1...$ as many as you want $*1$, the only restriction is the $4*1/2$ term, which means $n/2>=4+1/4 = 9$.
So for now, we only actually need to check 1, 2, 3, 5, 7, and 1 and 2 is obviously not possible.
So the remaining question is if 3, 5, 7 can be represented as the form mentioned in the question?
A: Not an answer (a supplement to the OP's answer). Notice that by the AGM (just as in the NNP's deleted answer):
$$n \geq k n^{\frac1k},$$
$$n^{1-\frac1k} \geq k.$$
Now, if $n=3,$ then
$$3^{1-\frac1k}\geq k.$$ The LHS is at most 2, and $\sqrt{3} < 2,$ so an impossibility.
For $5,$ $$5^{1-\frac1k} \geq k.$$ So, $k$ can only be equal to $2,$ and it is easy to check that this is impossible.$
For $7,$$ $$7^{1-\frac1k}\geq k,$ which means that $$7 \geq k^{k/(k-1)},$$ which means $k=2, 3, 4.$ $2$ is clearly impossible, but $3, 4$ are not obviously so.
