symmetric $(2k)\times (2k)$ matrix, $k \geq 4$, consisting of $0$s and $1$s with exactly $k$ ones in each row and column Let $k \geq 4$. Let $M=(m_{ij})_{1 \leq i,j\leq 2k}$ be a symmetric $(2k)\times (2k)$ matrix whose entries are either $0$s or $1$s and 
(1) All diagonal entries are zero, i.e., $m_{ii}=0$;
(2) Every row and every column have precisely $k$ ones.
(3) The total number of $1$s is $2k^2$.
(4) Assume that 
$$M \neq  \begin{bmatrix}0^{k\times k} &1^{k\times k}\\1^{k\times k} &0^{k\times k}\end{bmatrix},
$$
or any matrix that consists of an elementary permutation of this matrix (I could not think of a better way to say this).
Suppose that $M_1, \ldots, M_{2k}$ are the row vectors (also the column vectors by symmetry) of $M$. I want to prove the existence of two integers $1 \leq i<j \leq 2k$ such that $m_{ij}=m_{ji}=1$ and $M_i \cdot M_j^T\geq 2$. 
I cannot find a counterexample for smaller cases like $k=4$ and $k=5$. I wanted to try to prove this by contradiction: Suppose that for each $1\leq i<j\leq 2k$ such that $m_{ij}=m_{ji}=1$ and $M_i \cdot M_j^T\in \{0,1\}$. Here is where I hit a wall. Any suggestions?

I will include the graph theory equivalent problem: Let $k \geq 4$. Given a $k$-regular graph on $2k$ vertices and $k^2$ edges such that $G \not \cong K_{k,k}$, show that there exists an edge $xy \in E(G)$ such that $|N(x)\cap N(y)|\geq 2$, i.e., $x$ and $y$ have at least two neighbors. This is equivalent to showing that $G$ must contain a subgraph isomorphic to $K_4-e$. 
If it is of any help, there is a theorem by Paul Erdős that says that such a graph, as described above, should have at least $k-1$ triangles. I want to show that there exist two triangles that share an edge.

This--the statement you are intending to prove--is equivalent to saying that there are two adjacent vertices in any undirected $k$-regular graph $G$ on $2k$ vertices that isn't complete bipartite share at least 2 neighbours. Here is a proof for $k > 3$:
Let $v$ be an arbitrary vertex in $G$. Let $N(v)$ be the set of vertices of distance 1 (this set has $k$ vertices) and let $N_2(v)$ be the remaining $k-1$ vertices. By the fact that $G$ isn't complete bipartite there are at least two vertices $u$ and $w$ in $N(w)$ that are adjacent, lest every vertex in $N(v)$ be adjacent to all $k-1$ vertices in $N_2(v)$ [due to $k$-regularity]
Now let $C_w$ be the number of vertices in $N(v)$ that $w$ is adjacent to besides $u$ and let $C_u$ be the number of vertices in $N(v)$ that $u$ is adjacent to besides $w$. Then both $C_w$ and $C_u$ must be 0, or (assuming WLOG that $C_w$ is positive) $w$ is adjacent to $v$ and then would share two neighbors with $v$ (namely $u$ and the $C_w$ other vertices in $N(v)$).
On the other hand, $u$ is adjacent to $k-2-C_u$ vertices in $N_2(v)$ while $w$ is adjacent to $k-2-C_w$ vertices in $N_2(v)$. So for these sets to not overlap (if they overlap then we are done) then $2k-4-(C_u+C_w) \leq k-1$
implying $C_u+C_w \geq k-3$. As we already have shown in the previous paragraph that $C_u+C_w$ must be 0, this leads to a contradiction.
