We have given a prime $p$ s.t. $p^3 + 2p^2 + p = F(p)$ has exactly 42 divisors. Hence, since $F(p) = p(p + 1)^2$ and $p$ has two divisors, the perfect square $(p + 1)^2$ has ${\textstyle \frac{42}{2}=21}$ divisors (implying $p \neq 2$). Assume $\prod_{i=1}^k p_i^{r_i}$ is the prime factorization of $p+1$. Consequently $\prod_{i=1}^k (2r_i + 1) = 3 \cdot 7$, yielding $k=2$ and $(r_1,r_2)=(1,3), (3,1)$. This combined with the fact that $p+1$ is even (since $p$ is an odd prime) give us $p = 2q^3-1$ or $p=8q-1$, where $q$ is an odd prime. Conclusion:. The smallest prime $p$ for which $p^3 + 2p^2 + p$ has exactly 42 divisors, is $p = 8 \cdot 3 - 1 = 24 - 1 = 23$.