9.3. For a natural number $n$, denote by $T(n)$ the product of all its natural divisors (including $n$). a) Calculate $T(2022)$. b) Does there exist a prime number $p$ and a natural number $n$ such that $T(n)=p^{2022}$?
Answer: a) $2022^{4}$; b) does not exist. Solution. a) Factorize 2022 into prime factors: $2022=2 \cdot 3 \cdot 337$. Each even divisor of this number has the form $2 \cdot 3^{a} \cdot 337^{b}$, where the exponents $a, b$ take two values 0 or 1. The number of such ordered pairs $(a, b)$ is $4=2 \cdot 2$. Therefore, there are exactly 4 even divisors, and thus the factor 2 will enter the product in the fourth power. Similarly, the other prime divisors 3 and 337 will enter the product in the fourth power. Thus, $T(2022)=2^{4} \cdot 3^{4} \cdot 337^{4}=(2022)^{4}$. b) Since $p$ is a prime number, and $T(n)$ is a power of the number $p$, then $n$ does not have prime divisors other than $p$. Let $n=p^{y}$ for some natural number $y$. Then $T\left(p^{y}\right)=1 \cdot p^{1} \cdot p^{2} \cdot \ldots \cdot p^{y}=p^{1+2+\ldots+y}$. Therefore,
$(1+2+3+\ldots+y)=2022$. Thus, it is required to solve the quadratic equation $\frac{y(y+1)}{2}=2022 \Leftrightarrow y^{2}+y-4044=0$ in natural numbers. The discriminant of the obtained quadratic equation is 16177, which is not a perfect square (since it ends in 7). Therefore, the sought $n$ does not exist.