Determine for what $n\ge 3$ integer numbers, it is possible to find positive integer numbers $a_1 < a_2 < ...< a_n$ such $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$ and $a_1 a_2\cdot\cdot\cdot a_n$ is a perfect square.
Problem
Source: 2019 Peru MO (ONEM) L3 p1 - finals
Tags: number theory, Perfect Squares, Perfect Square
iniffur
26.11.2022 02:56
Hint
Starting from perfect numbers $6$ and $28$, it can be seen that
$6\Longrightarrow \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1 ,~~~ 1*2*3*6=6^2$
$28\Longrightarrow\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28}=1,~~~~ 1*2*4*7*14*28=28^3$
i.e., that the sum of the reciprocals of the divisors of a perfect number (omitting 1) $= 1$
However, for having a square, the number of factors (including 1 and the number itself) should be a multiple of 4.
This only occurs for $6$, not for $28, 496, 8128$ and $33550336$ which do not lead to a square.
pi_quadrat_sechstel
26.11.2022 13:33
parmenides51 wrote: Determine for what $n\ge 3$ integer numbers, it is possible to find positive integer numbers $a_1 < a_2 < ...< a_n$ such $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$ and $a_1 a_2\cdot\cdot\cdot a_n$ is a perfect square. For $n=3$ we can choose $(a_1,a_2,a_3)=(2,3,6)$ and for $n=4$ we can choose $(a_1,a_2,a_3,a_4)=(2,4,6,12)$. If the claim holds for $n$ then for $n+2$ by replacing $a_n$ with $2a_n,3a_n,6a_n$. So it works for every positive integer $n\geq3$.