NMO Selection Tests for the Junior Balkan Mathematical Olympiad - First Selection Test

Let $n$ be the required minimum value, and $a, b, c \in \mathbb{N}_{\ge 1}$ pairwise distinct, such that $n = a + b + c$ and $(a+b)(b+c)(c+a) = k^3$, where $k \in \mathbb{N}$.

If $a, b, c$ are all odd, then $n \ge 1+3+5=9$. If $n=9$, then $\{a, b, c\} = \{1, 3, 5\}$, which leads to $(a+b)(b+c)(c+a) = 4 \cdot 6 \cdot 8 = 192$, which isn't a perfect square. Consequently, $n \ge 11$.

Suppose now that not all the numbers $a, b, c$ have the same parity. WLOG, suppose that $a$ and $b$ have the same parity. It follows that $a+b$ is even and $c+a$ and $b+c$ are both odd. Then $2 \mid (a+b)(b+c)(c+a) = k^3$, so $2 \mid k$, which implies $8 \mid (a+b)(b+c)(c+a)$. Since $b+c$ and $c+a$ are odd, it follows that $8 \mid a+b$, so $a+b \ge 8$.
If $n=9$, then $a+b=8$ and $c=1$, and also $(b+c)(c+a) = ab+9$ must be a perfect cube, so $ab \ge 18$. Since $4 = \frac{a+b}{2} \ge \sqrt{ab}$, it follows that $ab \le 16$, a contradiction.

Consequently, $n \ge 10$. Notice that for $n=10$ we may consider $a=1, b=2, c=7$, for which $(a+b)(b+c)(c+a) = 216 = 6^3$, so the required minimum value is $n=10$.