In any group of 17 people, where each person knows 4 others, you can find 2 people, which don't know each other and have no common friends. I have a problem with proof of this (graph theory):

In any group of 17 people, where each person knows exactly 4 people, you can 
  find 2 people, which don't know each other and have no common friends.

I translated this to proving, that there exists a pair of vertices $\{v,w\}$, which aren't connected, that is, there isn't edge $(v,w)$ and for any other vertex $x$ from $V$ applies $(x, v) \veebar (x, w)$ or there is no edges between $x$ and $v$ and between $x$ and $w$, but then I am stuck. 
I tried using Pigeonhole Principle, but I couldn't use it correctly, I think. I couldn't use Ramsey theory too.
Any help and hints would be appreciated.
I drew two examples of these graphs for help:


 A: Let $G(V,E)$ be a $4$-regular simple graph on $17$ vertices.  We claim that there are two vertices $u,v\in V$ such that $u\neq v$, $u$ is not adjacent to $v$, and $u$ and $v$ do not have a common neighbor.  We prove by contradiction by assuming that, for any two distinct vertices $u$ and $v$ of $G$, if $u$ is not adjacent to $v$, then $u$ and $v$ have a common neighbor.  
Let $S$ denote all pairs $\big(\{u,v\},w\big)$ with $u,v,w\in V$ such that $u\neq v$, $\{u,v\}\notin E$, and $w$ is adjacent to both $u$ and $v$.  For each $w\in V$, $w$ has four neighbors.  Therefore, at most $\binom{4}{2}$ pairs of neighbors of $w$ are not adjacent.  This proves that
$$|S|\leq \binom{4}{2}\,|V|=6\,|V|=102\,.\tag{*}$$
Now, $|E|=\dfrac{4\cdot |V|}{2}=2\,|V|=34$ by the Handshake Lemma.  Thus, there are $$\binom{17}{2}-|E|=102$$ pairs of vertices $\{u,v\}$ that are not edges of $G$.  Each anti-edge pair $\{u,v\}$ produces at least one element of $S$, due to our hypothesis on $G$.  This proves that $$|S|\geq 102\,.\tag{#}$$
From (*) and (#), we must have $|S|=102$.  For (#) to be an equality, every anti-edge pair $\{u,v\}$ must have exactly one common neighbor $w\in V$.  Additionally, $G$ must be a triangle-free graph for (*) to become an equality.  This means $G$ is both triangle-free and quadrilateral-free.  Therefore, $G$ is a graph on $17=4^2+1$ vertices with girth $g\geq 5$ in which all vertices have degree $4$.  By the Hoffman-Singleton Theorem (for a proof, see here), if there exists an $r$-regular simple graph on $r^2+1$ vertices with girth at least $5$, then $r\in\{1,2,3,7,57\}$ (we know a partial converse, that is, for $r\in\{1,2,3,7\}$, there exists such a graph, but it is still a mystery for $r=57$, as you may guess, it is not easy to construct a graph on $57^2+1=3250$ vertices).  This yields the desired contradiction.
A: EDIT: I submitted a less than helpful response the first time. Here is my proof in this edit.

When all $17$ people within a group know $4$ people from the group, then there are $34$ friend pairings. 
In the above diagram, ensuring that everyone is at least a friend of a friend requires $52$ pairings "so far" just for persons $1$ through $5$. I have only worked the requirement for persons $1$ to $5$ because it already exceeds the requirement of knowing exactly $4$ others.
Every person in the group doesn’t personally know $12$ others in the group. But for there to be a possibility of sharing a friend with all $12$ others, every person’s $4$ friends must between them know all the other $12$.
In the diagram above, person $1$ personally knows $4$ others $(2,3,4,5)$. And between this $4$, they know all the other $12$ people ($6$ through $17$). But the same situation must exist for the friends of $1$, ($2,3,4$ and $5$). So, on the chart this requirement has been filled in where each set of $4$ friends for $2,3,4,5$ must know their corresponding other $12$. When this is done however, the number of friends for some of the people exceeds $4$. Not only that, but ensuring everyone is a least a friend of a friend hasn't been done for all $17$ in the group.
These are the $5$ acquaintance pairings so far:
$17 (8,11,12,13,14); 12 (6,7,8,10,17); 7(9,12,13,14,15)$
$16 (8,9,10,11,14); 11 (6,13,15,16,17); 6 (9,10,11,12,15)$
$15 (6,7,8,11,14); 10 (6,12,13,14,16)$ 
$14 (7,10,15,16,17); 9 (6,7,8,13,16)$
$13 (7,9,10,11,17); 8 (9,12,15,16,17)$
Therefore, for all unacquainted people to share a common friend, the unacquainted people have to know more than $4$ people. Hence with each person only knowing $4$ others, there will always be at least two people who don’t know each other and do not share a common friend.
A follow up question could be, what is the least number of  acquaintances each person must have to ensure that everyone is at least a friend of a friend?

