HMMT November, 2019

Solution:

Let $p_{1}, p_{2}, \ldots, p_{k}$ be the prime divisors of $n$. Then it is known that $\varphi(n)=n \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}}$. As $n^{2}$ and $n$ have the same set of prime divisors, it also holds that $\varphi\left(n^{2}\right)=n^{2} \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}}$. We will examine the equality in four cases.

- $\operatorname{gcd}(n, 2019)=1$ In this case, $2019 \cdot n$ has also $3$ and $673$ as prime divisors, thus $\varphi(2019 \cdot n)=2019 \cdot n \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}} \cdot \frac{2}{3} \cdot \frac{672}{673}$, and the equality implies $n=1342$, however $\operatorname{gcd}(1342,3) \neq 1$, contradiction. Thus, there is no answer in this case.

- $\operatorname{gcd}(n, 2019)=3$ In this case, $2019 \cdot n$ has also $673$ as a prime divisor, thus $\varphi(2019 \cdot n)=2019 \cdot n \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}} \cdot \frac{672}{673}$, and the equality implies $n=2016$, which satisfies the equation. Thus, the only answer in this case is $n=2016$.

- $\operatorname{gcd}(n, 2019)=673$ In this case, $2019 \cdot n$ has also $3$ as a prime divisor, thus $\varphi(2019 \cdot n)=2019 \cdot n \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}} \cdot \frac{2}{3}$, and the equality implies $n=1346$, which satisfies the equation. Thus, the only answer in this case is $n=1346$.

- $\operatorname{gcd}(n, 2019)=2019$ In this case, $2019 \cdot n$ has the same set of prime divisors, thus $\varphi(2019 \cdot n)=2019 \cdot n \cdot \frac{p_{1}-1}{p_{1}} \ldots \frac{p_{k}-1}{p_{k}}$, and the equality implies $n=2019$, which satisfies the equation. Thus, the only answer in this case is $n=2019$.

Thus, all the answers are $n=1346, 2016, 2019$.