TITLE: Twelve Distinct Positive Integers
QUESTION [11 upvotes]: Let S be a set of twelve distinct positive integers such that for distinct a, b, c, and d in S, 
a + b ≠ c + d. Show that the largest element in S is greater than 56.
I found some math competition problems, and I am trying to work my way through them, but I am encountering difficulties. With this one, I am thinking that somehow it needs to be shown that if the largest number is ≤56 then we can find four distinct such that a + b = c + d. Then we have a contradiction.

REPLY [3 votes]: Consider the ${12 \choose 2} = 66$ possible differences $b-a$, where $b>a$ are the elements
of $S$.
Note that if $a, b, c, d \in S$ are pairwise distinct and $a-c = d-b$, then $a+b = c+d$.
Also, if
$a,b,c \in S$ are all distinct and $c-a = b-c$, then $c = (a+b)/2$ so $c$ is not the largest or smallest element of $S$. 
Next, if $c-a = b -c$ and $c-d = e-c$, then $a+b = 2c = d+e$.  So for
each $c$, that is not the largest or smallest element of $S$, there exists at most one pair of $a, b$ that can be found such that $c-a=b-c$. Because if not, then we will have $a + b = d +e$.
Hence, there are at most $10$ duplicates for each $c$ that is not the largest or
smallest element of $S$.  Therefore, there are at least
$66 - 10 = 56$ distinct differences. 
Hence, the largest element of $S$ is greater than 56.