2013 OIM Problems/Problem 5

Problem

Let $A$ and $B$ be two sets such that:

1. $A \cup B$ is the set of positive integers.

2. $A \cap B$ is empty.

3. If two positive integers have as difference a prime greater than 2013, then one of them is in $A$ and the other in $B$ .

Find all the possibilities for sets $A$ and $B$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

We know that $2027$ and $2029$ are primes greater than $2013$ by testing. As a result, let some number $x$ be in $A$ . Then $x+2029$ is in $B$ , so $(x+2029)-2027=x+2$ is in $A$ . We can also perform this in the opposite direction; given some $x\in A$ , $x+2027$ is in $B$ , so $(x+2027)-2029=x-2$ is in $A$ .

We can repeat this process, and we can easily prove using induction that the set $A$ contains all of the positive integers $x+2k$ for integers $k$ ; therefore, $A$ must contain either only the even integers or only the odd integers. Clearly, $B$ must then contain only the odd integers or only the even integers, respectively. Both of these cases work (because in order for two positive integers to have a difference as such a prime, they must differ by an odd integer, and separating odds and evens facilitates this). As a result, the two solutions are $\boxed{A=\{x|x\equiv0\pmod{2}\},B=\{x|x\equiv1\pmod{2}\}}$ $\text{or}$ $\boxed{A=\{x|x\equiv1\pmod{2}\},B=\{x|x\equiv0\pmod{2}\}}$ ~ eevee9406

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2013 OIM Problems/Problem 5

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