Vietnamese MO

a) First, we will prove the following lemma.

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

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 $s\le \sqrt{n}$.

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 $\tau (n)\le 2s\le 2\sqrt{n}$. $■$

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 $n\le 2\sqrt{n}+2023$. Solving this inequality, we get $$n\le (\sqrt{2024}+1{)}^2<2115.$$ 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 $$(e_1+1)(e_2+1)\cdots (e_k+1)+2023=p_1^{e_1}{p}_2^{e_2}\cdots p_k^{e_k}(1)$$ 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 $3\le p_1<p_2<\cdots <p_k$.

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 $$2\tau (n)+2023=n.(2)$$ 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 $$2(e_1+1)(e_2+1)\cdots (e_s+1)+2023=p_1^{e_1}{p}_2^{e_2}\cdots p_s^{e_s}.(3)$$ From equation (3), we deduce that $p_1,p_2,\dots ,p_s$ are odd prime numbers. Without loss of generality, assume that $3\le p_1<p_2<\cdots <p_s$. 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. $\square$