Unique common friend given a group of $m$ people 
Let $m \geq 3$ be a fixed integer. Given $n$ people in a room, we know
that any group of $m$ people in the room have a unique common friend.
Express $n$ in terms of $m.$ (Assume that friendship is mutual, i.e. if A is a friend of B then B is friends with A.)


First, I treated the people as points on a graph and let friendships be edges. Then, I started experimenting with some small cases and I noticed that in the case of $n = m+1,$ the complete graph $K_n$ satisfied the given conditions.
However, I am unsure if there are more possible $n.$
 A: Throughout this answer I'll be using the graph terminology you introduced (vertices are people and edges represent friendship). We will use induction to show that for all $m\geq 3$ the only graph with the property that every $m$ vertices have exactly one common neighbor is the complete graph $K_{m+1}$ and therefore prove that $n=m+1$.

*

*For $m=3$ consider a graph $G$ with that property and let $x,y,z \in V(G)$ be three of its vertices. From our hypothesis there exists a common neighbor $v$ of $x,y,z$ in $G$. Clearly $|N(v)| \geq 3$ and let $H=G[N(v)]$ be the subgraph induced by the neighbors of $v$. Now every two vertices in $H$ have exactly one common neighbor also in $H$. We claim that $|N(v)|=3$. Suppose that $|N(v)|>3$. Then, from the friendship theorem, there exists a vertex in $H$ that is connected to all other vertices in $H$. If $a,b,c$ are three vertices of $H$ and $d$ is their common neighbor in $H$ then in the initial graph $G$, the vertices $a,b,c$ have two common neighbors, namely $v$ and $d$, a contradiction. Therefore, $|N(v)|=3$ and $H$ has to be a triangle. Since we picked $x,y,z$ arbitrarily and showed that they are pairwise connected, $G$ has to be a complete graph. The only valid choice for $G$ is $K_4$ and the base case is proven.


*For the inductive step consider a graph $G$ with the property that every $m$ vertices have exactly one common neighbor and let $S \subseteq V(G)$ be a subset of its vertices of size $m$. Again, let $v$ be the common neighbor of $S$ and let $H=G[N(v)]$. Now every $m-1$ vertices in $H$ have exactly one common neighbor in $H$ so from the inductive hypothesis $H=K_m$ and using the same reasoning as above $G$ has to be a complete graph, so $G=K_{m+1}$.
As a sidenote, I'm not sure this is the best approach since it uses the friendship theorem which is a famous yet non-trivial result in graph theory. Perhaps there are other approaches that avoid it.
