If $n\geqslant 3$ is an integer and $a_1,a_2,\dotsc ,a_n$ are non-zero integers such that $$a_1a_2\cdots a_n\left( \frac{1}{a_1^2}+\frac{1}{a_2^2} +\cdots +\frac{1}{a_n^2}\right)$$is an integer, does it follow that the product $a_1a_2\cdots a_n$ is divisible by each $a_i^2$?
Problem
Source: Stars of Mathematics 2019, Senior, P2
Tags: number theory
ThE-dArK-lOrD
01.12.2019 09:20
Yes.
Suppose to the contrary that $\nu_p(a_i)>\sum_{j\neq i}{\nu_p(a_j)}$ for some prime $p$.
We easily get that $\nu_p$ of the expression is less than zero.
microsoft_office_word
23.03.2021 18:11
Let $b_i = \dfrac{a_1a_2 \ldots a_n}{a^2_i}$. We take the polynomial $(X - b_1)(X-b_2) \ldots (X-b_n)$. This polynomial has degree $n$, is monic and has $n$ rational roots $b_1, b_2 \ldots b_n$. If we prove that all coefficents of the polynomial are integers, we are done, because in this case all $b_1, b_2 \ldots b_n$ must be integers. We are given that the coefficient of $X^{n-1}$ is an integer, and for other coefficents, the power of $a_i$ in the numerator is $\ge 2$ and the power of $a_i$ in the denominator is either $0$ or $2$. Hence $a^2_i$ should divide $a_1a_2 \ldots a_n$, for each $1\le i \le n$.