Given a positive integer $n$, define $\tau(n)$ as the number of positive divisors of $n$ and $\sigma(n)$ as the sum of those divisors. For example, $\tau(12) = 6$ and $\sigma(12) = 28$. Find all positive integers $n$ that satisfy: \[ \sigma(n) = \tau(n) \cdot \lceil \sqrt{n} \rceil \]
Problem
Source: Brazilian TST 1 P3 Cono Sur/OMCPLP 2024
Tags: number theory, Divisors, Sigma function, tau function, ceiling function
HoshimiyaMukuro
17.10.2024 13:41
\begin{align*} \tau (n)\cdot \lceil\sqrt{n}\rceil=\sigma (n)&=1+n+\sum_{d|n,d\ne 1,n}d \\&=1+n+\frac{1}{2}\sum_{d|n,d\ne 1,n}(d+\frac{n}{d}) \\&\ge 1+n+\frac{1}{2}\sum_{d|n,d\ne 1,n}\lceil 2\sqrt{n}\rceil \\&\ge 1+n+\sum_{d|n,d\ne 1,n}\lceil \sqrt{n}\rceil \\&= 1+n+(\tau (n)-2)\lceil \sqrt{n}\rceil \end{align*}So we have $1+n\le 2\lceil\sqrt{n}\rceil<2\sqrt{n}+2\Rightarrow (\sqrt{n}-1)^{2}<2$,then it's easy to get $n=1,3,5$.
Thelink_20
26.10.2024 16:18
Let $1=d_1<d_2<\dots<d_{\tau (n)}=n$ be the divisors of $n$. By $AM\geq GM, \ d_i+d_{\tau(n)-i+1}\geq 2\sqrt{n}\implies \sigma(n)\geq [\tau(n)-2]\sqrt{n}+(n+1)$.
There is a bijection between the divisors less than $\sqrt{n}$ and the ones larger. Because there are at most $\sqrt{n}$ divisors less than $\sqrt{n}$, it follows. (If $n$ is a square it's the same).
. Now we will show that $\sigma(n)>\tau(n)\cdot\lceil \sqrt{n}\rceil \ $ for $n\geq 16$. It suffices to show that: $\sigma(n)\geq\tau(n)\cdot (\sqrt{n}+1) \Leftarrow [\tau(n)-2]\sqrt{n}+(n+1)\geq\tau(n)\cdot (\sqrt{n}+1)\Leftarrow$ $\Leftarrow n-2\sqrt{n}+1\geq\tau(n)\Leftarrow n-2\sqrt{n}+1\geq 2\sqrt{n}$, which is true for $\sqrt{n}\geq 4$. Testing yields $n=1, 3, 5, 6 \ _{\blacksquare}$
lksb
16.12.2024 22:18
Thelink_20 wrote: Let $1=d_1<d_2<\dots<d_{\tau (n)}=n$ be the divisors of $n$. By $AM\geq GM, \ d_i+d_{\tau(n)-i+1}\geq 2\sqrt{n}\implies \sigma(n)\geq [\tau(n)-2]\sqrt{n}+(n+1)$.
There is a bijection between the divisors less than $\sqrt{n}$ and the ones larger. Because there are at most $\sqrt{n}$ divisors less than $\sqrt{n}$, it follows. (If $n$ is a square it's the same).
. Now we will show that $\sigma(n)>\tau(n)\cdot\lceil \sqrt{n}\rceil \ $ for $n\geq 16$. It suffices to show that: $\sigma(n)\geq\tau(n)\cdot (\sqrt{n}+1) \Leftarrow [\tau(n)-2]\sqrt{n}+(n+1)\geq\tau(n)\cdot (\sqrt{n}+1)\Leftarrow$ $\Leftarrow n-2\sqrt{n}+1\geq\tau(n)\Leftarrow n-2\sqrt{n}+1\geq 2\sqrt{n}$, which is true for $\sqrt{n}\geq 4$. Testing yields $n=1, 3, 6 \ _{\blacksquare}$ 6?