Determine all squarefree positive integers $n\geq 2$ such that \[\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_k}\]is a positive integer, where $d_1,d_2,\ldots,d_k$ are all the positive divisors of $n$.
Problem
Source: Romania JBMO TST 2022
Tags: number theory, Romanian TST, squarefree
CT17
21.04.2022 01:06
Notice that since $n$ is squarefree, $\sum_{d|n} \frac{1}{d} = \prod_{p | n}\frac{p+1}{p}$ (where the product is over primes $p$). Suppose that $n$ is divisible by a prime $p_0\ge 5$. Then to cancel the denominator of $\frac{p_0+1}{p_0}$, another prime $p_1\ge 2p_0 - 1$ must divide $n$. Using this reasoning, we can construct an infinite sequence of prime divisors of $n$ such that $p_{i+1}\ge 2p_i - 1$ for all $i$. This is absurd, so we have a contradiction. Hence, all prime divisors of $n$ are in the set $\{2,3\}$, and checking yields that only $n=6$ works.
Iora
04.05.2022 20:59
My solution is different than @above , so I am sharing it
Determine all squarefree positive integers $n\geq 2$ such that \[\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_k}\]is a positive integer, where $d_1,d_2,\ldots,d_k$ are all the positive divisors of $n$.
If $d_i$ is a divisor of $n$, then so is $\frac{n}{d_i}$, and $\frac{1}{\frac{n}{d_i}}= \frac{d_i}{n}$, Therefore we have:
$$ \sum_{i=1}^k \frac{d_i}{n}= \frac{d_1+d_2+...+d_k}{n}$$
Note that all of the $d_i$ are square free and $n| \sigma (n)$ where $\sigma(n)$ is sum of all factors of squarefree number. Then $n \le \sigma(n)$, note that $\sigma(n)$ is always even, hence $n$ is even. Let:
$$n=2 \cdot p_2 \cdot ... \cdot p_k$$
Then $\sigma(n)= (1+2)(1+p_2)(1+p_3)...(1+p_k)$,
Now let $p_k$ be the greatest prime divisor of $n$ other than $2$. Then $ p_k| p_j+1$ for some $j$. Then $p_k \le p_j+1$. But $p_k$ is the greatest prime, There can't be 2 neighbour primes other than $2,3$, and contraction, hence the only primes are $2,3$, $n=6$