SELECTION TESTS FOR THE 2019 JBMO

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 $n^3+m_1+n^3+m_2+\cdots +n^3+m_b=k(n^3+n_1+n^3+n_2+\cdots +n^3+n_a)$. Then $n^3(ka-b)=m_1+m_2+\cdots +m_b-k(n_1+n_2+\cdots +n_a)$.

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 $n^3+1+n^3+2+\cdots +n^3+n\ge n^3+m_1+n^3+m_2+\cdots +n^3+m_b\ge (n+1)(n^3+n_1+n^3+n_2+\cdots +n^3+n_a)\ge (n+1)n^3$, 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 $m_1+m_2+\cdots +m_b-k(n_1+n_2+\cdots +n_a)<1+2+\cdots +n=\frac{n(n+1)}{2}<n^3$ and $m_1+m_2+\cdots +m_b-k(n_1+n_2+\cdots +n_a)\ge -n(n_1+n_2+\cdots +n_a)\ge -n(1+2+\cdots +n)=-\frac{n^2(n+1)}{2}>-n^3$.

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