
Problem 4. Given a positive integer $n$, denote by $\tau(n)$ the number of positive divisors of $n$, and by $\sigma(n)$ the sum of all positive divisors of $n$. Find all positive integers $n$ satisfying

$$
\sigma(n)=\tau(n) \cdot\lceil\sqrt{n}\rceil
$$

(Here, $\lceil x\rceil$ denotes the smallest integer not less than $x$.)

(Michael Reitmeir)


Solution. Answer: $n \in\{1,3,5,6\}$.

We consider two cases:

(1) $n$ is a square: $n=1$ is a solution, so assume $n>1$. Squares have an odd number of positive divisors, so we write $\tau(n)=2 k+1$ with $k>0$ and $d_{k+1}=\sqrt{n}$. Furthermore, $\lceil\sqrt{n}\rceil=\sqrt{n}$, so the given equation is equivalent to

$$
d_{1}+\cdots+d_{k}+d_{k+2}+\cdots+d_{2 k+1}=2 k \cdot \sqrt{n}
$$

For $i \in\{1, \ldots, k\}, d_{i}$ and $d_{2 k+2-i}$ are complementary divisors, i.e. $d_{i}$. $d_{2 k+2-i}=n$. Thus, we obtain an equivalent equation:

$$
\begin{array}{r}
\left(d_{1}-2 \sqrt{n}+d_{2 k+1}\right)+\left(d_{2}-2 \sqrt{n}+d_{2 k}\right)+\cdots+\left(d_{k}-2 \sqrt{n}+d_{k+2}\right)=0 \\
\Longleftrightarrow \quad\left(\sqrt{d_{1}}-\sqrt{d_{2 k+1}}\right)^{2}+\left(\sqrt{d_{2}}-\sqrt{d_{2 k}}\right)^{2}+\cdots+\left(\sqrt{d_{k}}-\sqrt{d_{k+2}}\right)^{2}=0 .
\end{array}
$$

Since squares are non-negative and $d_{1}, \ldots, d_{2 k+1}$ are pairwise distinct, the equation has no solutions.

(2) $n$ is not a square: Then no divisor of $n$ is its own complementary divisor, so $n$ has an even number of positive divisors and we write $\tau(n)=2 k$. We shall prove the following inequality for $n$ sufficiently large:

$$
d_{1}+\cdots+d_{2 k}>2 k \cdot\lceil\sqrt{n}\rceil
$$

First, note that $\sqrt{n}+1 \geq\lceil\sqrt{n}\rceil$, so it suffices to show

$$
d_{1}+\cdots+d_{2 k}>2 k \cdot(\sqrt{n}+1) .
$$

For $i \in\{1, \ldots, i\}, d_{i}$ and $d_{2 k+1-i}$ are complementary divisors, so $d_{i} \cdot d_{2 k+1-i}=$ $n$. Hence we obtain an equivalent inequality:

$$
\begin{aligned}
&\left(d_{1}-2 \sqrt{n}+d_{2 k}\right)+\left(d_{2}-2 \sqrt{n}+d_{2 k-1}\right)+\cdots+\left(d_{k}-2 \sqrt{n}+d_{k+1}\right)>2 k \\
& \Longleftrightarrow \quad\left(\sqrt{d_{1}}-\sqrt{d_{2 k}}\right)^{2}+\left(\sqrt{d_{2}}-\sqrt{d_{2 k-1}}\right)^{2}+\cdots+\left(\sqrt{d_{k}}-\sqrt{d_{k+1}}\right)^{2}>2 k .
\end{aligned}
$$

But for $n$ sufficiently large, the term $\left(\sqrt{d_{1}}-\sqrt{d_{2 k}}\right)^{2}=(\sqrt{n}-1)^{2}$ is already bigger than the right-hand side: Indeed, the numbers $\left\lfloor\frac{n}{2}\right\rfloor+1,\left\lfloor\frac{n}{2}\right\rfloor+2, \ldots, n-$ 1 do not divide $n$, so $2 k=\tau(n) \leq\left\lfloor\frac{n}{2}\right\rfloor+2 \leq \frac{n}{2}+2$. As $\frac{n}{2}$ grows faster than $2 \sqrt{n}$ for large $n$, we will get the desired estimate. Indeed, this happens for $n \geq 20$, since

$$
\begin{aligned}
& \frac{n}{2}=\frac{\sqrt{n}}{2} \cdot \sqrt{n} \geq \frac{\sqrt{20}}{2} \cdot \sqrt{n}=2 \sqrt{n}+(\sqrt{5}-2) \sqrt{n} \geq \\
& \quad \geq 2 \sqrt{n}+(\sqrt{5}-2) \sqrt{20}=2 \sqrt{n}+10-2 \sqrt{5}>2 \sqrt{n}+1
\end{aligned}
$$

the last inequality resulting from $\sqrt{5}2 k \cdot s$, thus it remains to check $n \in\{2,3,5,6,7\}$ of which only $n \in\{3,5,6\}$ work.

Pairing the divisors $d_{1}, \ldots, d_{2 k}$ up such that product in each pair is constant, we get $d_{2 k}+d_{1}>d_{2 k-1}+d_{2}>\cdots>d_{k+1}+d_{k}$, where each sum is an integer. Moreover, even the smallest sum satisfies $d_{k+1}+d_{k}>2 \sqrt{n}>2(s-1)$, and since both sides are integers we get $d_{k+1}+d_{k} \geq 2 s-1$. It remains to show that for $n \geq 8$ we have $n+1=d_{2 k}+d_{1}>2 s+1$, i.e. $n>2 s$. This is simple algebra: it suffices to show $n>2(\sqrt{n}+1)$, i.e. $(\sqrt{n}-1)^{2}>3$, which is true for $n>(\sqrt{3}+1)^{2}=4+2 \sqrt{3} \doteq 7.5$. (As a slightly weaker bound, for $n \geq 9$ (thus $\sqrt{n} \geq 3$ ) we have $n \geq 2 \sqrt{n}+\sqrt{n}>$ $2(\sqrt{n}+1)>2 s)$.
