2. Find the smallest prime number $p$ for which the number $p^{3}+2 p^{2}+p$ has exactly 42 positive divisors.
III/2. First, we write $p^{3}+2 p^{2}+p=p(p+1)^{2}$. Since $p$ and $p+1$ have no common divisors (except 1), each divisor of the number $p(p+1)^{2}$ is either 1 times some divisor of the number $(p+1)^{2}$ or $p$ times that divisor. Since the number $p(p+1)^{2}$ has exactly 42 divisors, $(p+1)^{2}$ has exactly 21 divisors. From the prime factorization

$$
(p+1)^{2}=\left(p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}\right)^{2}=p_{1}^{2 \alpha_{1}} \cdot p_{2}^{2 \alpha_{2}} \cdots p_{k}^{2 \alpha_{k}}
$$

we determine that $(p+1)^{2}$ has exactly $\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{k}+1\right)$ divisors. This means that $2 \alpha_{1}+1=3$ and $2 \alpha_{2}+1=7$ and $\alpha_{i}=0$ for $i>2$. Therefore, $(p+1)^{2}=p_{1}^{2} \cdot p_{2}^{6}$ or $p+1=p_{1} \cdot p_{2}^{3}$. From this, we get $p=p_{1} \cdot p_{2}^{3}-1$. The smallest value is obtained if we choose $p_{1}=3$ and $p_{2}=2$. Then, indeed, $p=3 \cdot 8-1=23$, which is a prime number.