Maybe : https://fr.wikipedia.org/wiki/Graphe_eul%C3%A9rien

tenplusten wrote: In the country some mathematicians know each other and any division of them into two sets contain 2 friends from different sets.It is known that if you put any set of four or more mathematicians at a round table so that any two neighbours know each other , then at the table there are two friends not sitting next to each other.We denote by $c_i $ the number of sets of $i$ pairwise familiar mathematicians(by saying "familiar" it means know each other).Prove that $c_1-c_2+c_3-c_4+...=1$ Translated into graphs language it reads like: A graph $G$ is given satisfying the following properties: (i) Any cycle with length at least $4$ contains a chord. (ii) $G$ is connected. Prove that $c_1-c_2+c_3-c_4+...=1$. Let $f(G) := c_1-c_2+c_3-c_4+...$ We will follow the following plan. Starting from a graph $G$ that satisfies $(i)$ and $(ii)$ we will add consecutively new edges to $G$ keeping intact $(i)$ and $(ii)$, till we end up with a complete graph. We will check $f(G)$ doesn't change at every step. Since $f(G)=1$ for any complete graph, it would imply $f(G)=1$ for any graph satisfying $(i)$ and $(ii)$. So, staring from $G$, we add such an edge that would bring to arising additional maximal complete graph (clique) inside $G$. Let it be the edge $v_1\,v_2$. We denote the new graph as $G'$ and that max clique containing $v_1,v_2$ be $Q\subset G'$. Apparently $G'$ also satisfies $(ii)$. Suppose $C := v_1\,v_2\,v_3\dots v_n\,v_1$ is a cycle with length $n\geq 4$. We want to prove there exists an edge $v_i\,v_j$ for some non adjacent $v_i, v_j$ inside $C$. Suppose on the contrary there are no other edges between the vertices of $C$ except those of the cycle itself. Then $v_3,\dots,v_n$ are outside the clique $Q$. Playing with the property $(i)$ which holds for the graph $G$ , we can conclude the vertices $v_3,\dots,v_n$ are connected to all the vertices of $Q\setminus \{v_1,v_2\}$. It means that $(Q\setminus v_1)\cup \{v_3,v_4\}$ would become a clique if we add the edge $v_2\,v_4$. But it contradicts with the maximality of the clique $Q$ because of the choice of $v_1\,v_2$. Thus, we've shown $G'$ also satisfies $(i), (ii)$. Now, let's check $f(G)=f(G')$. Consider $f(G')-f(G)= \Delta c_1-\Delta c_2+\Delta c_3-\dots$ , where $\Delta c_i$ is the number of the additional $i$-cliques arisen after adding the edge $v_1\,v_2$. Let us show all that additional cliques are inside that maximal clique $Q$. Assume there exists a clique $Q'$ containing $v_1,v_2$ and not included into $Q$. Playing again with property $(i)$, we can conclude all vertices of $Q$ and $Q'$ are connected, which contradicts with the maximality of $Q$. Thus, $\Delta c_i = \binom{n-2}{i-2} \,,\,i=2,3,\dots,n$ (here $n=|Q|$), hence $-\Delta c_2+\Delta c_3-\dots = 0$. Since $\Delta c_1=0$ (no new vertices), we get $f(G)=f(G')$. The result follows.

dgrozev wrote: A graph $G$ is given satisfying the following properties: (i) Any cycle with length at least $4$ contains a chord. (ii) $G$ is connected. Prove that $c_1-c_2+c_3-c_4+...=1$. $\textbf{My solution :}$ Let $f(G) = c_1(G)-c_2(G)+\ldots$. Induction on $|G|$. Consider any vertex $v\in G$. So one get that $f(G) = 1 + f(G\setminus\{v\}) - f(G_v\setminus\{v\})$, where $G_v$ is a subgraph of $G$ consisting of all vertexes $u\in G$, such that there is an edge between $v, u$. Here we will consider only the case when $G\setminus\{v\}$ is connected, general case is similar. So from induction we get that $1 + f(G\setminus\{v\})=2$, so it's enough to show that $f(G_v\setminus\{v\})=1$. By induction we only need to check that $G_v\setminus\{v\}$ is connected. And it easily follows from the connectivity of $G\setminus\{v\}$ and a condition $(i)$. Done

So, you take a graph $G$ satisfying (i),(ii), then fix some vertex $v$ and $G_v$ is the subgraph induced by the vertices of $G$ that $v$ is connected to. But then $G_v$ may not be connected! Of course, one can consider its connected components, but you should do some calculations, it's not so obvious. I tried that approach at first, but it got too complicated. Maybe I'm wrong, but it couldn't be carried out like that, at least if you choose $v$ arbitrary.

Yes, $v$ is any and $G_v$ is neighbors of $v$. We know that from $(i)$, $G_v\setminus\{v\}$ is connected iff $G\setminus \{v\}$ is. And connected components of $G_v\setminus\{v\}$ are exactly connected components of $G\setminus\{v\}$ induced on $G_v\setminus\{v\}$. From this observations and from induction we get that $f(G_v\setminus\{v\}) = f(G\setminus\{v\})$ and $f(G) = 1 +f(G\setminus\{v\}) - f(G_v\setminus\{v\}) = 1$.

@Mosquitall: It seems you're right. Apparently I miscalculated something when considering that the most natural approach.

Consider the simplicial complex $X$ with the vertex set being the set of mathematicians, and the simplex formed by a subset of mathematicians are in $X$ iff every pair of them are friends. The first condition implies that $X$ is connected, the second condition implies that $X$ is contractible. The conclusion follows since the Euler characteristic of $X$, viz the left side of the equation, by the homotopy invariance of homology is the same as that of a point, viz 1.

So, staring from $G$, we add such an edge that would bring to arising additional maximal complete graph (clique) inside $G$. Is it obvious that we can add such edge?