Vietnamese MO

a) First, we will prove the following lemma.

Lemma. For any positive integer $n$, then $\tau(n) \leq 2\sqrt{n}$.

Proof. Consider any positive integer $n$, let $d_1, d_2, \dots, d_s$ be all positive divisors not exceeding $\sqrt{n}$ of $n$. Then, obviously we have $s \leq \sqrt{n}$.

Note that, if $x$ is a divisor not less than $\sqrt{n}$ of $n$ then $\frac{n}{x}$ is a positive divisor not exceeding $\sqrt{n}$ of $n$. It follows that $\tau(n) \leq 2s \leq 2\sqrt{n}$. $\blacksquare$

Back to the problem, suppose there exists a positive integer $n$ satisfying the equation $\tau(n) + 2023 = n$. According to the above lemma, we have $n \leq 2\sqrt{n} + 2023$. Solving this inequality, we get
$$n \leq (\sqrt{2024} + 1)^2 < 2115.$$
On the other hand, we also have $n = \tau(n) + 2023 \ge 2025$. Therefore $2025 \le n \le 2114$. Let $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ where $p_1, p_2, \dots, p_k$ are distinct primes and $e_1, e_2, \dots, e_k$ are positive integers. Then $\tau(n) = (e_1 + 1)(e_2 + 1)\cdots(e_k + 1)$ and the equation $\tau(n) + 2023 = n$ can be rewritten as
$$(e_1 + 1)(e_2 + 1)\cdots (e_k + 1) + 2023 = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \quad (1)$$
If $e_1, e_2, \dots, e_k$ are all even numbers, then $n$ is a perfect square, and $2025 \le n \le 2114$ so $n = 2025$. However, when try again, this value does not satisfy the equation. Therefore, among the numbers $e_1, e_2, \dots, e_k$ there must be at least one odd number. From here, combined with equation (1), we deduce that $p_1, p_2, \dots, p_k$ are odd prime numbers. Without loss of generality, assume $3 \le p_1 < p_2 < \dots < p_k$.

If $k \ge 5$, then we have $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \ge 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 > 2114$, a contradiction. Therefore $k \le 4$.

Case 1: $k = 1$. In this case, we have $2114 \ge n = p_1^{e_1} \ge 3^{e_1}$. Therefore $e_1 \le 6$. Checking each case of the value of $e_1$, we find no prime number $p_1$ that satisfies the equation (1).

Case 2: $k = 2$. In this case, we have $2114 \ge n = p_1^{e_1} p_2^{e_2} \ge 3^{e_1} 5^{e_2} \ge 3^{e_1+e_2-1} \cdot 5$, deduce $e_1 + e_2 \le 6$. From there
$$\tau(n) = (e_1 + 1)(e_2 + 1) \le \left( \frac{e_1 + e_2 + 2}{2} \right)^2 \le 9.$$
Checking specific cases of $\tau(n)$, we find no corresponding value of $n$ satisfying the equation.

Case 3: $k = 3$. In this case, we have
$$2114 \ge n = p_1^{e_1} p_2^{e_2} p_3^{e_3} \ge 3^{e_1} 5^{e_2} 7^{e_3} \ge 3^{e_1+e_2+e_3-2} \cdot 5 \cdot 7,$$
infers $e_1 + e_2 + e_3 \le 5$. Which $e_1 + e_2 + e_3 \ge 3$ so $e_1 + e_2 + e_3 \in \{3, 4, 5\}$, infers $(e_1, e_2, e_3)$ is a permutation of one of the sets of numbers $(1, 1, 1)$, $(1, 1, 2)$, $(1, 1, 3)$, $(1, 2, 2)$. From there $\tau(n) \in \{8, 12, 16, 18\}$. Checking all the specific cases of $\tau(n)$, we find no corresponding value of $n$ which satisfies the equation.

Case 4: $k = 4$. In this case, we have $2114 \ge n = p_1^{e_1} p_2^{e_2} p_3^{e_3} p_4^{e_4} \ge 3^{e_1} 5^{e_2} 7^{e_3} 11^{e_4}$. If $e_1, e_2, e_3, e_4$ has a large number than 1, then $3^{e_1} 5^{e_2} 7^{e_3} 11^{e_4} \ge 3^2 \cdot 5 \cdot 7 \cdot 11 > 2114$, which is a contradiction. Therefore $e_1 = e_2 = e_3 = e_4 = 1$, i.e. $\tau(n) = (e_1+1)(e_2+1)(e_3+1)(e_4+1) = 16$. However, when substituting back to the equation $\tau(n)+2023 = n$, we find no corresponding value $n$ which satisfies.

So, the equation $\tau(n)+2023 = n$ has no positive integer solution.

b) All prime numbers $k > 6996$ satisfy the problem requirements. Indeed, considering any prime number $k$, greater than 6996. We have the following comment
$$\tau(kn) \le 2\tau(n).$$
Indeed, let $n = k^e \cdot p_1^{e_1} p_2^{e_2} \cdots p_s^{e_s}$, where $p_1, p_2, \dots, p_s$ is Other distinct prime numbers $k$; $e_1, e_2, \dots, e_s$ are positive integers and $e$ is a natural number. Then, we have
$$\tau(kn) = (e+2)(e_1+1)(e_2+1)\cdots(e_s+1) \le 2(e+1)(e_1+1)(e_2+1)\cdots (e_s+1) = 2\tau(n).$$
In addition, we also see that, if $n$ is not divisible by $k$ (i.e. $e = 0$), then $\tau(kn) = 2\tau(n)$.

Now, consider a positive integer solution $n$ (if any) of the equation $\tau(kn) + 2023 = n$. Using the above result, combined with the Lemma in part a), we have
$$n = \tau(kn) + 2023 \le 2\tau(n) + 2023 \le 4\sqrt{n} + 2023.$$
Solving the inequality $n \le 4\sqrt{n}+2023$, we get $n \le (2 + \sqrt{2027})^2 < 6996 < k$. It follows that $n$ is not divisible by $k$, from which $\tau(kn) = 2\tau(n)$ and the equation $\tau(kn)+2023 = n$ is rewritten as
$$2\tau(n) + 2023 = n. \qquad (2)$$
We will prove that equation (2) has exactly two solutions: $n = 2027$ and $n = 2031$. Indeed, above, we have proved
$$n \le (2 + \sqrt{2027})^2 < 2212.$$
Furthermore, from (2), we have $n = 2\tau(n)+2023 \ge 2027$. Therefore $2027 \le n \le 2211$. Let $n = p_1^{e_1} p_2^{e_2} \cdots p_s^{e_s}$ where $p_1, p_2, \dots, p_s$ are distinct primes and $e_1, e_2, \dots, e_s$ are positive integers. Then $\tau(n) = (e_1+1)(e_2+1)\cdots(e_s+1)$ and equation (2) can be rewritten as
$$2(e_1 + 1)(e_2 + 1)\cdots (e_s + 1) + 2023 = p_1^{e_1} p_2^{e_2} \cdots p_s^{e_s}. \quad (3)$$
From equation (3), we deduce that $p_1, p_2, \dots, p_s$ are odd prime numbers. Without loss of generality, assume that $3 \le p_1 < p_2 < \dots < p_s$. At this point, we note that, if $s \ge 5$ then
$$2211 \ge n = p_1^{e_1} p_2^{e_2} \cdots p_s^{e_s} \ge 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 > 2211,$$
which is a contradiction. Therefore $s \le 4$. Consider the following cases.

Case 1: $s = 1$. In this case, we have $2211 \ge n = p_1^{e_1} \ge 3^{e_1}$. Therefore $e_1 \le 7$. Checking all values of $e_1$, we find exactly one solution $n = 2027$ which satisfies.

Case 2: $s = 2$. In this case, we have
$$2211 \ge n = p_1^{e_1} p_2^{e_2} \ge 3^{e_1} 5^{e_2} \ge 3^{e_1+e_2-1} \cdot 5,$$
so $e_1 + e_2 \le 6$. It follows that
$$\tau(n) = (e_1 + 1)(e_2 + 1) \le \left( \frac{e_1 + e_2 + 2}{2} \right)^2 \le 9.$$
Checking all cases of $\tau(n)$, we find exactly one value $n = 2031$ which satisfies.

Case 3: $s = 3$. In this case, we have
$$2211 \ge n = p_1^{e_1} p_2^{e_2} p_3^{e_3} \ge 3^{e_1} 5^{e_2} 7^{e_3} \ge 3^{e_1+e_2+e_3-2} \cdot 5 \cdot 7,$$
therefore $e_1 + e_2 + e_3 \le 5$. So $e_1 + e_2 + e_3 \in \{3, 4, 5\}$, therefore $(e_1, e_2, e_3)$ is a permutation of one of the following triplets $(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 2)$. We deduce that $\tau(n) \in \{8, 12, 16, 18\}$. Checking all cases of $\tau(n)$, we find no corresponding value $n$ which satisfies.

Case 4: $s = 4$. In this case, we have
$$2211 \ge n = p_1^{e_1} p_2^{e_2} p_3^{e_3} p_4^{e_4} \ge 3^{e_1} 5^{e_2} 7^{e_3} 11^{e_4}.$$
If one of $e_1, e_2, e_3, e_4$ is greater than 1, then $3^{e_1} 5^{e_2} 7^{e_3} 11^{e_4} \ge 3^2 \cdot 5 \cdot 7 \cdot 11 > 2211$, which is a contradiction. Therefore $e_1 = e_2 = e_3 = e_4 = 1$, i.e. $\tau(n) = (e_1 + 1)(e_2 + 1)(e_3 + 1)(e_4 + 1) = 16$. However, when substituting back to equation (2), we find no corresponding value of $n$ which satisfies.

Thus, the equation (2) has exactly two positive integer solutions: $n = 2027$ and $n = 2031$. It follows that the equation $\tau(kn) + 2023 = n$ has exactly two positive integer solutions: $n = 2027$ and $n = 2031$. Hence the problem is proved. $\square$