sitting around a table- graph theory? $50$ mathematicians attend a conference at which each knows $25$ other attendees. Show that you can select $4$ of them who can then be seated at a round table, such that each person at the table knows the two people he or she is sitting next to.

Should I use graph theory to show this?
 A: Construct a graph with 50 vertices, each having degree 25 (each edge is a relationship). By the handshake lemma, there are 625 edges.
The problem asks you to prove that such a graph contains a 4-circuit.
Thm: If $e \gt \frac{n}{4}(1+ \sqrt{4n-3})$ then G has a 4-circuit, where e is the number of edges, and n is the number of vertices  in G.
Proof: Assume a G has no 4-circuits. Now we will count pairs of edges with a vertex in common in 2 ways, call this number P.
First we know that 2 vertices can have at most 1 pair of edges adjacent to them such that the edges meet in another vertex, so $P \le \binom{n}{2}$.
The number of pairs of this sort can also be counted from the perspective of each vertex, 
$$P = \sum_{i=1}^{n} \binom{deg(v_i)}{2} = \frac{1}{2} \sum deg(v_i)^2 - deg(v_i)$$
by Cauchy-Schwarz and the handshake lemma,  $$ \sum deg(v_i)^2 \ge \frac{1}{n} (\sum deg(v_i))^2 = \frac{1}{n} (2e)^2 $$
Therefore, 
$$ \frac{1}{2} (n^2-n) \ge P \ge \frac{1}{2} (\frac{4e^2}{n} - 2e) $$
Solving the resulting quadratic inequality for e yields the desired result.
