Call a positive integer $n$ good if for every relatively prime positive integers $a$ and $b$ that divides $n$, $a + b - 1 | n$. Finding all good positive integersWe claim that $n$ is a good positive integer if and only if either $n = p^k$ for some prime number $p$ and positive integer $k$, or $n = 1$, or $n = 12$. It is easy to check that all such positive integers are good, so we are left to prove that they are the only good positive integers. Suppose that $n$ has $m \geq 2$ prime divisors, and let $n = \prod_{i = 1}^m p_i^{k_i}$ be the prime factorization of $n$, where $p_1 < p_2 < \ldots < p_m$. Let $\prod_{i = 2}^m p_i^{k_i} = T > 1$ (so $n = p_1^{k_1} T$ and $\gcd(p, T) = 1$). First, consider $a = p_1$ and $b = T$. Then, $p_1 + T - 1$ is a divisor of $n$. In particular, the set of all prime divisors of $p_1 + T - 1$ s a subset of $\{p_1, p_2, \ldots, p_k\}$. But for $i\geq 2$, $p_1 + T - 1 \equiv p_1 - 1 \not\equiv 0 \pmod{p_i}$, so $p_1$ is the only possible prime divisor of $p_1 + T - 1$. In particular, since $p_1 + T - 1 > p_1$, then $2 \leq v_{p_1}(p_1 + T - 1) \leq v_{p_1}(n) = k_1$. Next, consider $a = p_1^2$ and $b = T$. This time, as $p_1^2 | p_1 + T - 1$, we have $v_p(p_1^2 + T - 1) = 1$. Also, $p_1^2 + T - 1 | n = p_1^{k_1} T$, so $\frac{p_1^2 + T - 1}{p_1} | T$. Now, assume for the sake of contradiction that $\frac{p_1^2 + T - 1}{p_1}$ is not prime. Since it is greater than 1, that means there exists not necessarily distinct indices $i$ and $j$ with $2\leq i, j\leq m$ such that $p_i p_j | \frac{p_1^2 + T - 1}{p_1}$; i.e. $p_i p_j | p_1^2 + T - 1$. But $p_i p_j > p_1^2 - 1 > 0$ and $p_i p_j | \frac{p_1^2 + T - 1}{p_1} | T$, which leads to a contradiction. Hence, $\frac{p_1^2 + T - 1}{p_1}$ is a prime number, and thus it equals $p_i$ for some $2\leq i\leq m$. If $T$ is composite, then $T \geq p_i p_2 > p_i p_1$ and thus $\frac{p_1^2 + T - 1}{p_1} > p_i$; a contradiction. Hence, $T = \prod_{i = 2}^m p_i^{k_i}$ is prime. In particular, $m = 2$ and $k_2 = 1$, implying $n = p_1^{k_1} p_2$. From the above result as well, $\frac{p_1^2 + p_2 - 1}{p_1} = p_2$, so $p_2 = p_1 + 1$. On the other hand, $p_1 + p_2 - 1 = p_1^c$ for some $c \geq 2$, so $2 p_1 = p_1^c$, implying $p_1 = 2$, and thus $p_2 = 3$. So, $n = 3 \cdot 2^{k_1}$ for some $k_1\geq 2$. If $k_1\geq 3$, then picking $a = 3$ and $b = 8$, we have $10 | n$; a contradiction. So, $k_1 = 2$ and thus $n = 12$. Hence, the only good positive integer with at least 2 prime factors is $12$. Finishing partCall a pair of positive integers $(x, y)$ an excellent pair if $x^3 + y^3$ is good. The previous result means that $x^3 + y^3 = p^k$ for some prime number $p$ and positive integer $k$, as there is no pair of positive integers $x$ and $y$ such that $x^3 + y^3 \in \{1, 12\}$. Observe that if $(x, y)$ is an excellent pair, then so is $\left( \frac{x}{\gcd(x, y)}, \frac{y}{\gcd(x, y)}\right)$, so first we will find all excellent pairs $(x, y)$ with $\gcd(x, y) = 1$. Meanwhile, if $(x, y)$ is an excellent pair, then $x + y$ is a power of a prime number $p$ and $(xz, yz)$ is an excellent pair if and only if $z$ is also a power of $p$. Now, observe that $x + y = p^t$ and $x^2 - xy + y^2 = p^u$ for some prime number $p$, positive integer $t$ and non-negative integer $u$. Furthermore, $\gcd(x + y, x^2 - xy + y^2) = \gcd(x + y, 3y) = \gcd(x + y, 3)$, so if $p \neq 3$, then $\gcd(x + y, x^2 - xy + y^2) = 1$. In particular, this yields $x^2 - xy + y^2 = 1$ and thus $x = y = 1$. Otherwise, if $p = 3$, then since $(x, y)\neq (1, 1)$, we have $x^2 - xy + y^2 > 1$, so either $x + y = 3$ or $x^2 - xy + y^2 = 3$; but it is easy to check either way that both yields $(x, y) \in \{(2, 1), (1, 2)\}$. Generalizing, we found that all excellent pairs are of form $(2^k, 2^k)$, $(2\cdot 3^k, 3^k)$, and $(3^k, 2\cdot 3^k)$ for some non-negative integer $k$.