TITLE: 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.
QUESTION [9 upvotes]: 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:

REPLY [2 votes]: Here is an extremely tedious combinatorial proof...  :P  It works for this case but most likely cannot generalize.
Lemma 1: The following are taken from the first part of the answer by @Batominovski

The graph $G$ is triangle-free and quadrilateral-free.
If $(u,v)$ is not an edge then $u, v$ have exactly 1 common neighbor, which we will denote $N(u,v)$.

Let the nodes be $\{1, 2, ..., 17\}$.  The following tables show successive "forced" constructions of the adjacency matrix in "sparse format", i.e. a row "a: b, c, d, e" means there are edges a-b, a-c, a-d and a-e.
WOLOG we can pick node 1 and choose its 4 neighbors and their neighbors, to start:
 1:  2  3  4  5
 2:  1  6  7  8
 3:  1  9 10 11
 4:  1 12 13 14
 5:  1 15 16 17

Now consider nodes $\{6,7,...,17\}$ organized into 4 triplets $T_i: i \in \{2,3,4,5\}$ where each $T_i =$ the 3 nodes (among 6...17) connected to $i$.  E.g. $T_3 = \{9,10,11\}$.
Lemma 2: For $i \neq j: \{N(u,j): u \in T_i\} = T_j$, i.e. the function $N(.,j)$ is bijective from $T_i$ to $T_j$.
Proof: first of all $N(u,j) \in T_j$ because there is no other way to connect to $j$.  (The only other neighbor of $j$ is $1$ which does not connect to $u$.)  Next $u \neq u' \implies N(u,j) \neq N(u',j)$ because otherwise $(N(u,j), u, i, u', N(u',j))$ would be a quadrilateral.
Based on Lemma 2, WOLOG we can fill out the neighbors of every $u \in T_2$:
 6:  2  9 12 15
 7:  2 10 13 16
 8:  2 11 14 17

At this point we arrange the remaining nodes 9...17 and visualize them in a 3x3 grid:
   9   10   11  .. [3]
  12   13   14  .. [4]
  15   16   17  .. [5]
   :    :    :
  [6]  [7]  [8]

The 3 nodes of any given row have a common neighbor (3, 4 or 5, in []) and the 3 nodes of any given column also have a common neighbor (6, 7, 8, in []).
Node 13 still has 2 edges left.  It cannot connect to 12/14 on the same row (13-12/14-4-13 triangle) nor 10/16 on the same column.  The 2 edges must also connect to different columns (e.g. 13-9 & 13-15 would lead to 13-9-6-15-13 quadrilateral) and different rows.  WOLOG we connect 9-13 and 13-17.
   9   10   11  .. [3]
     \
  12   13   14  .. [4]
          \
  15   16   17  .. [5]
   :    :    :
  [6]  [7]  [8]

Now after connecting 9-13, node 9 has 1 edge left.  This cannot connect to 10/11 (same row as 9), nor 12/15 (same column), nor 16 (9-16-7-13-9 quad) nor 14 (9-14-4-13-9 quad), so it must connect to 17.  But this forms a triangle 9-13-17-9, a contradiction.
Whew...  and yuck.  :P