European Mathematical Cup

Let $f(x,y)=x^m{y}^n-x^n{y}^m$. If $n=m$, the problem statement will be fulfilled no matter how we choose $B$ so from now on, without loss of generality, we consider $n>m$. Since $m$ and $n$ are both odd, we have that $n-m$ is even and we get $$\begin{array}{rl}f(x,y)& =x^m{y}^m(y^{n-m}-x^{n-m})\\ f(x,y)& =x^m{y}^m(y^2-x^2)Q(x,y)\\ f(x,y)& =x^m{y}^m(y-x)(y+x)Q(x,y)\end{array}$$ where $Q(x,y)=y^{n-m-2}+y^{n-m-4}x+\cdots +y^{n-m-2}$.

Now if one of $x,y$ is even, $f(x,y)$ is even. If both are odd, then $f(x,y)$ is again even since $x+y$ and $x-y$ are even in that case. This shows that we only need to consider divisibility by $5$.

If $A$ contains at least one element divisible by $5$, we can put it in $B$ and that will give us the solution easily.

Now we consider the case when none of the elements in $A$ is divisible by $5$. If some two numbers in $A$ give the same remainder modulo $5$, we can choose them and then $x-y$ will be divisible by $5$ which solves the problem.

Now we consider the case when all remainders modulo $5$ in $A$ are different. Take a look at the pairs $(1,4)$ and $(2,3)$. Since we have three different remainders modulo $5$, by pigeonhole principle one of these pairs has to be completely in $A$ (when elements are considered modulo $5$). Then if we pick the numbers from $A$ that correspond to those two remainders we get that $x+y$ is divisible by $5$ so the problem statement is fulfilled again. This completes the proof.