A combinatorics problem Given $A = \{a_0, a_1,...,a_m\}$  such that it's a subset of $\{1,2,...,n\}$ where $m>n/2$, and $a_0$ is the smallest number in $A$. Show that $A$ contains two numbers $b$ and $c$ such that $a_0+b=c$ with the use of the pigeonhole principle. 
Consider the following if stuck: $a_1-a_0, a_2-a_0,...,a_m-a_0$
Hi, I am new to this site. Please help, I have been stuck on this problem for hours now. How would you go about it? 

Hint: Consider the set $S = \{a_1 - a_0, a_2 - a_0, \ldots, a_m - a_0\}$. If an element in $S$ is also in $A$, then this would mean that $a_p - a_0 = a_q$, for some $p, q$. Rearranging, you would get $a_0 + a_q = a_p$, so we can take $b = a_q$ and $c = a_p$.
Now the question is: Is there an element in $S$ that is also in $A$?
