Problem 10.6. A pair of natural numbers ( $a, p$ ) is called good if the number $a^{3}+p^{3}$ is divisible by $a^{2}-p^{2}$, and $a>p$.

(a) (1 point) Indicate any possible value of $a$ for which the pair $(a, 13)$ is good.

(b) (3 points) Find the number of good pairs for which $p$ is a prime number less than 20.
Answer: (a) any of the numbers $14, 26, 182$. (b) 24.

Solution. Since $a^{3}+p^{3}=(a+p)\left(a^{2}-a p+p^{2}\right)$, and $a^{2}-p^{2}=(a+p)(a-p)$, the condition of divisibility is equivalent to $a^{2}-a p+p^{2}=a(a-p)+p^{2}$ being divisible by $a-p$. Note that $a(a-p)$ is divisible by $a-p$, so $p^{2}$ must be divisible by the natural number $a-p$, i.e., $a-p$ is a divisor of $p^{2}$. Since $p$ is a prime number, $p^{2}$ has only 3 natural divisors: $1, p$, and $p^{2}$. Therefore, for any prime $p$, there are exactly 3 options for the number $a$: these are $1+p, p+p$, and $p^{2}+p$ (obviously, each of these numbers is greater than $p$).

(a) For $p=13$, we get $a=14,26,182$.

(b) Let's find the number of prime numbers less than 20. There are exactly 8: $2,3,5,7,11,13,17,19$. Thus, the total number of good pairs is $3 \cdot 8=24$.