Selection Tests for the JBMO, 2019

Denote $A = \{n^3 + n_1, n^3 + n_2, \dots, n^3 + n_a\}$, $B = \{n^3 + m_1, n^3 + m_2, \dots, n^3 + m_b\}$ and $k \in \mathbb{N}$ such that $n^3 + m_1 + n^3 + m_2 + \dots + n^3 + m_b = k(n^3 + n_1 + n^3 + n_2 + \dots + n^3 + n_a)$. Then $n^3(ka - b) = m_1 + m_2 + \dots + m_b - k(n_1 + n_2 + \dots + n_a)$.

The case $n = 1$ is obviously true since $M = \{1, 2\}$ and we can only have $A = \{1\}$ and $B = \{2\}$.

Now let's prove that $n > 1$ implies $k < n + 1$ (so $k \le n$). Indeed, supposing $k \ge n + 1$, we would get $n^3 + 1 + n^3 + 2 + \dots + n^3 + n \ge n^3 + m_1 + n^3 + m_2 + \dots + n^3 + m_b \ge (n+1)(n^3 + n_1 + n^3 + n_2 + \dots + n^3 + n_a) \ge (n+1)n^3$, therefore $n^4 + \frac{n(n+1)}{2} \ge n^4 + n^3$. This would imply $n^2 + n \ge 2n^3$ which is false.

We are left with the case $k \le n$. Now we have $m_1 + m_2 + \dots + m_b - k(n_1 + n_2 + \dots + n_a) < 1 + 2 + \dots + n = \frac{n(n+1)}{2} < n^3$ and $m_1 + m_2 + \dots + m_b - k(n_1 + n_2 + \dots + n_a) \ge -n(n_1 + n_2 + \dots + n_a) \ge -n(1 + 2 + \dots + n) = -\frac{n^2(n+1)}{2} > -n^3$.

To summarize, we have the inequalities $-n^3 < n^3(ka - b) < n^3$, therefore $ka - b = 0$ showing that $a$ divides $b$.