Problem:
Find all positive integers n such that there are k≥2 positive rational numbers a1​,a2​,…,ak​ satisfying a1​+a2​+…+ak​=a1​⋅a2​⋯ak​=n.
Solution:
The answer is n=4 or n≥6.
I. First, we prove that each n∈{4,6,7,8,9,…} satisfies the condition.
(1). If n=2k≥4 is even ​, we set (a1​,a2​,…,ak​)=(k,2,1,…,1) :
a1​+a2​+…+ak​=k+2+1⋅(k−2)=2k=n
and
a1​⋅a2​⋅…⋅ak​=2k=n
(2). If n=2k+3≥9 is odd​, we set (a1​,a2​,…,ak​)=(k+23​,21​,4,1,…,1) :
(3). A very special case is n=7​, in which we set (a1​,a2​,a3​)=(34​,67​,29​). It is also easy to check that
a1​+a2​+a3​=a1​⋅a2​⋅a3​=7=n
II. Second, we prove by contradiction that each n∈{1,2,3,5} fails to satisfy the condition.
Suppose, on the contrary, that there is a set of k≥2 positive rational numbers whose sum and product are both n∈{1,2,3,5}. By the Arithmetic-Geometric Mean inequality, we have
This proves that none of the integers 1,2,3, or 5 can be represented as the sum and, at the same time, as the product of three or more positive numbers a1​,a2​,…,ak​, rational or irrational.
The remaining case k=2 also goes to a contradiction. Indeed, a1​+a2​=a1​a2​=n implies that n=a12​/(a1​−1) and thus a1​ satisfies the quadratic
a12​−na1​+n=0
Since a1​ is supposed to be rational, the discriminant n2−4n must be a perfect square (a square of a positive integer). However, it can be easily checked that this is not the case for any n∈{1,2,3,5}. This completes the proof.
Remark. Actually, among all positive integers only n=4 can be represented both as the sum and product of the same two rational numbers. Indeed, (n−3)2<n2−4n=(n−2)2−4<(n−2)2 whenever n≥5; and n2−4n<0 for n=1,2,3.