Problem 1767

1. Definitions and Initial Setup:
- Let $G$ be a finite simple graph with the property that each of its edges is contained in at most $n$ cycles.
- We need to prove that the chromatic number of $G$ is at most $n+1$.

2. Claim 1:
- Statement: $G$ has a $\chi$-critical $(n+2)$-chromatic subgraph $H$ with $V(H) = V(G)$.
- Proof:
- There exists a subgraph $G_0$ of $G$ with $\chi(G_0) = n+2$.
- If $G_0$ is $\chi$-critical, the claim is trivial.
- If not, there exists an edge $e \in E(G_0)$ such that $\chi(G_0 - e) = n+2$.
- Repeat this process until obtaining a $\chi$-critical $(n+2)$-chromatic subgraph $H$ of $G$. This process terminates because $G$ is finite. $\blacksquare$

3. Claim 2:
- Statement: Let $H$ be the $\chi$-critical $(n+2)$-chromatic graph. Then every pair of adjacent vertices can be connected by $n$ edge-disjoint even paths.
- Proof:
- Let $(x, y) \in E(H)$. Since $H$ is $\chi$-critical, $H - (x, y)$ has a coloring $\gamma$ with $n+1$ colors: $1, 2, \ldots, n+1$.
- Since $H$ is not $(n+1)$-colorable, assume without loss of generality that $\gamma(x) = \gamma(y) = 1$.
- Consider another color $j \in S := \{2, \ldots, n+1\}$ and the graph $H_j$ induced by the vertices of colors $1$ and $j$.
- Let $H_j'$ be the connected component of $H_j$ containing $x$. If $y \notin V(H_j')$, we can switch the colors $1$ and $j$ in $V(H_j')$, resulting in a $(n+1)$-coloring of $H$, a contradiction.
- Hence, $y \in V(H_j')$. Since $H_j'$ is connected, there exists a path $P_j$ from $x$ to $y$ in $H_j'$.
- Each edge of $P_j$ connects a vertex of color $1$ to a vertex of color $j$, making $P_j$ an even-length path.
- The paths $P_2, \ldots, P_{n+1}$ must be edge-disjoint. This proves the claim. $\blacksquare$

4. Claim 3:
- Statement: The graph $H$ (from Claim 1) contains an even cycle.
- Proof:
- Let $(a, b) \in E(H)$. From $n \geq 2$ and Claim 2, there are two edge-disjoint even paths $P$ and $Q$ from $a$ to $b$, forming the circuit $\mathcal{C} = P \cup Q$.
- If $\mathcal{C}$ is a cycle, we are done.
- Otherwise, let $M := (V(P) \cap V(Q)) \setminus \{a, b\} \neq \varnothing$. Let $v \in M$ be the vertex closest to $a$ on $P$.
- Define $P'$ as the path from $a$ to $v$ in $P$ and $P''$ as the path from $v$ to $b$ in $P$. Similarly, define $Q'$ and $Q''$.
- The cycle $\mathcal{C}' = P' \cup Q'$ intersects the circuit $P'' \cup Q''$ only at $v$.
- If $\mathcal{C}'$ is even, we are done. If $\mathcal{C}'$ is odd, then $1 \equiv |\mathcal{C}'| = |P'| + |Q'|$, implying $|P'| \not\equiv |Q'|$.
- Since $|P| \equiv 0$, it follows that $|Q' \cup P''| = |Q'| + |P''| \not\equiv |P'| + |P''| = |P| \equiv 0$, implying that the cycle formed by $Q' \cup P''$ and the edge $(a, b)$ is even. $\blacksquare$

5. Final Argument:
- Using Claim 3, consider an even cycle $\mathcal{C}$ in $H$ and $(x, y) \in E(\mathcal{C})$.
- From Claim 2, there are $n$ edge-disjoint even paths $Q_1, \ldots, Q_n$ from $x$ to $y$ in $H$.
- The cycles $\mathcal{C}_k = (V(Q_k), E(Q_k) \cup (x, y))$ are of odd lengths for $k = 1, \ldots, n$, implying at least $n+1$ cycles in $G$ passing through the edge $(x, y)$, contradicting the given condition that each edge is contained in at most $n$ cycles.

The final answer is $\boxed{ \chi(G) \leq n+1 }$.