Let $G$ be a undirected graph such that every two vertices have an odd number of common neighbors. Then the graph $G$ has a Euler circle.
It is straightforward to see that $G$ is connected as every two vertices have atleast one common neighbor.
I tried to approach the problem in the following manner - let $v_1,\ldots,v_n$ be the vertices of the graph, and let $u_i\in\mathbb{R}^n$ be the vector that is $1$ in the $j$'th spot if $(v_i,v_j)$ is an edge and $0$ otherwise (basically it is the $i$'th row of the adjacency matrix of $G$). By the assumption, for every $i\neq j$: $$\langle u_i,u_j\rangle$$ is odd, and there is an Euler circle iff all vertices have even degree, iff $\|u_i\|^2$ is even for all $i$.
I didn't know how to show this. I know there are different solution, but Im interested in knowing if there is a way to solve it using linear algebra as such. Any help would be appreciated.
HINT: For each $v \in G$, let $N(v)$ denote the set of vertices adjacent to $v$ in $G$. Then the condition that $N(v) \cap N(w)$ is odd implies that each $w \in N(v)$ is such that $w$ has an odd number of neighbors in $N(v)$. Thus the induced subgraph of $G$ on $N(v)$ has only vertices of odd degree. Thus, the Handshaking Lemma implies from this that $N(v)$ has to have an even number of vertices for each such $v$. And so $G$ must have an Eulerian cycle.
I think this can be put another way: Represent each vertex $v \in V(G)$ as a binary vector corresponding to its row in the adjacency matrix of $G$, so $v \cdot w$ is the number of common neighbors that $v$ and $w$ have in $G$, for any two vertices $v,w \in V(G)$. Then for each $v \in V(G)$ and $w \in N(v)$, the inner product $v \cdot w$ is $|\{w,u\}; u \in N(v) \cap N(w)|$. Thus, for each $v \in V(G)$:
$$\Big(\sum_{w \in N(v)} w \cdot v\Big) \ = \ 2\times \Big|\{w,u\} \in E(G); w, u \in N(v) \Big|,$$ and as the RHS of the above is even and $w \cdot v$ is odd for each such $v,w$ as above, it follows that the LHS must represent a sum of an even number of odd terms i.e., $N(v)$ must itself be even. [Note that each edge $\{w,u\} \in E(G)$; $u,w \in N(v)$; is counted twice as the edge $\{w,u\}$ contributes $1$ to each of both $v \cdot w$ and $v \cdot u$. Hence the factor of $2$ on the RHS.]