Contest Math: construct equations with$ \sum_{i=1}^k a_i = \prod_{j=1}^ka_j$ where $a_k$ are all positive rational number.

Question

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...

Answer 1

Apply the AM-GM inequality

$$n = \sum_{i=1}^na_n \ge n\sqrt[n]{\prod_{i=1}^na_i} =n\sqrt[n]{n} \iff n \ge n\sqrt[n]{n} \iff 1 \ge n$$

So, the necessary condition is $n=1$. And of course, this is also the sufficient condition.

Conclusion: $n = 1$ is the unique solution.

Answer 2

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?

Answer 3

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.