# Showing a pair exists whose sum or difference is even

I am looking at the following number theory problem: I am given any $m+2$ integers, and must show there exist two of them whose sum (or difference) is divisible by $2m$.

My first observation was that if any two of these integers are equal, their difference is 0 and hence trivially divisible by $2m$. Moreover, I take that for any two distinct integers, their difference will be divisible by $2m$ iff they have the same remainder (when divided by $2m$).

Then it should follow that given the proposition we wish to show, if the remainders of the $m+2$ integers are distinct, there exist two numbers whose sum (but not difference) is divisible by $2m$.

So to restate the problem, from a set of $m+2$ integers each with distinct remainders, how can I show there are two whose sum is divisible by $2m$?

If you look at your set of $m+2$ integers mod $2m$, you have only $2m$ distinct possible numbers. As you pointed out, if any are equal then their difference will be $0$. If the numbers are all distinct, then you can only have m numbers that won't add to $0$ mod $2m$ since $(2m - x) + x = 0$ $($mod $2m)$ since every number x where $0 \le x \le m-1$ there is a corresponding number of the form $(2m - x)$ where $m \le (2m-x) \le 2m$. From there you can say that that if there are more than $m$, (such as $m+2$) then there are two that add up to $0$ mod $2m$.
Showing that this works with a difference after showing it works with a sum is pretty easy, just show that there is a number of the form $(2m - x)$ mod $2m$ that is the negative of one of the numbers in your set.