Given a natural number $n$, we define $\tau(n)$ as the number of natural numbers that divide $n$, and we define $\sigma(n)$ as the sum of these divisors. Find all natural numbers $n$ for which

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

For a real number $x$, the notation $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$.
Answer: this only applies to 1, 3, 5, 6.
For these four cases, it indeed holds respectively that $1=1 \cdot 1, 4=2 \cdot\lceil\sqrt{3}\rceil, 6=2 \cdot\lceil\sqrt{5}\rceil$ and $12=4 \cdot\lceil\sqrt{6}\rceil$. From now on, we assume that $n \neq 1$.

If $n$ is a square, then $n$ has an odd number of divisors $1=d_{1}\sqrt[2 k-1]{\prod_{i=1}^{2 k-1} d_{i}}=\sqrt[2 k-1]{\sqrt{n} \cdot n^{k-1}}=\sqrt{n}=\lceil\sqrt{n}\rceil,
$$

since we cannot achieve equality because $d_{1} \neq d_{2 k-1}$. For $n$ being a square, there are thus no solutions.

If $n$ is not a square, then $n$ has an even number of divisors. Indeed, if $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{t}^{e_{t}}$ then $\tau(n)=\left(e_{1}+1\right)\left(e_{2}+1\right) \cdots\left(e_{t}+1\right)$ and at least one of the $e_{i}$ must be odd. We order the divisors again by size $1=d_{1}e+\frac{n}{e}$. Note that this is a strict inequality because both sequences are strictly increasing. With $e=d_{k}$ we find that $d+\frac{n}{d}>d_{k}+d_{k+1} \geq 2\lceil\sqrt{n}\rceil-1$. Thus $d_{k-i}+d_{k+1+i}>2\lceil\sqrt{n}\rceil-1$, or $d_{k-i}+d_{k+1+i} \geq 2\lceil\sqrt{n}\rceil$ for all $1 \leq i \leq k-1$.

Suppose now that $n \geq 8$. Then we know that $(n-4)^{2} \geq 4^{2}>12$, or $n^{2}-8 n+16 \geq 12$. This can also be written as $n^{2}-4 n+4 \geq 4 n$ and then factored as $(n-2)^{2} \geq 4 n$. Taking the square root of this, we find that

$$
\frac{n-1}{2} \geq \frac{n-2}{2} \geq \sqrt{n}
$$

Just as before, we note that $\frac{n-1}{2}$ is now greater than or equal to a natural number greater than or equal to $\sqrt{n}$, so $\frac{n-1}{2} \geq\lceil\sqrt{n}\rceil$. This can be rewritten as $n+1 \geq 2\lceil\sqrt{n}\rceil+2$.

For $n \geq 8$ and not a square, we conclude that

$$
\begin{aligned}
\sigma(n) & =\left(d_{k}+d_{k+1}\right)+\left(d_{k-1}+d_{k+2}\right)+\ldots+\left(d_{2}+d_{2 k-1}\right)+\left(d_{1}+d_{2 k}\right) \\
& \geq(2\lceil\sqrt{n}\rceil-1)+2\lceil\sqrt{n}\rceil+\ldots+2\lceil\sqrt{n}\rceil+(2\lceil\sqrt{n}\rceil+2) \\
& =2 k\lceil\sqrt{n}\rceil+1 \\
& =\tau(n) \cdot\lceil\sqrt{n}\rceil+1,
\end{aligned}
$$

which means there can be no solutions. It remains to check that for $n=2$ it holds that $3 \neq 2 \cdot\lceil\sqrt{2}\rceil$ and that for $n=7$ it holds that $8 \neq 2 \cdot\lceil\sqrt{7}\rceil$. The case $n=4$ was handled when we ruled out the squares.