Proof with pigeonhole principle

Question

I have a set $A={1,3,5,7,...,153,155}$, and $S\subset A$ so that $|S|=22$.

I have to prove that it is possible to choose 4 different subsets of size 2 in S, so that they have the same absolute difference.

Answer 1

There are $231$ subsets of size $2$ in $S$. The difference between the two elements of a pair is an even number between $2$ and $154$, so there are $77$ possibilities.

Suppose for a contradiction that for each of the $77$ possibilities for the difference there are at most $3$ subsets of $S$ of size $2$ with that difference. Then there would be exactly $3$ subsets of size $2$ with that difference, since $231 = 77\cdot 3$. So there are three subsets with difference $154$. But this is a contradiction, since the only subset with difference $154$ is $\{155,1\}$.