Prove that every k-regular graph with girth 4 has at least 2k vertices.

To prove that every \( k \)-regular graph with girth 4 has at least \( 2k \) vertices, we need to utilize the definitions and properties of \( k \)-regular graphs and girth.

### Definitions:
1. **\( k \)-regular graph:** A graph where each vertex has exactly \( k \) edges.
2. **Girth of a graph:** The length of the shortest cycle in the graph.

### Proof:

1. **Assume \( G \) is a \( k \)-regular graph with girth 4.**

2. **Girth 4 implies the smallest cycle in \( G \) has length 4.** Thus, there are no cycles of length 3 or less in \( G \).

3. **Consider a vertex \( v \) in \( G \).** Since \( G \) is \( k \)-regular, \( v \) has exactly \( k \) neighbors.

4. **Let \( N(v) \) be the set of neighbors of \( v \).** Therefore, \(|N(v)| = k \).

5. **Vertices in \( N(v) \) must form an independent set:** 
   - If any two vertices in \( N(v) \) were adjacent, there would be a cycle of length 3 involving \( v \), contradicting the girth of 4. Therefore, no two vertices in \( N(v) \) can be adjacent.

6. **Next step is to consider the neighbors of each vertex in \( N(v) \):**
   - Each vertex \( u \in N(v) \) has \( k-1 \) neighbors other than \( v \) since \( G \) is \( k \)-regular and \( v \) is already one neighbor.
   - Let's denote the set of neighbors of a vertex \( u \in N(v) \), excluding \( v \), as \( N(u) \setminus \{v\} \).

7. **The vertices in \( \bigcup_{u \in N(v)} (N(u) \setminus \{v\}) \) must be distinct:**
   - Consider any two vertices \( u, w \in N(v) \). If there were a common neighbor between \( u \) and \( w \), it would form a cycle of length 3 or create multiple 4-cycles with shared edges, which are not allowed by the definition of girth 4.
   - Therefore, \( (N(u) \setminus \{v\}) \cap (N(w) \setminus \{v\}) = \emptyset \) for any distinct \( u, w \in N(v) \).

8. **Counting vertices:**
   - The set \( N(v) \) has \( k \) vertices.
   - Each vertex in \( N(v) \) has \( k-1 \) additional neighbors, making \( k(k-1) \) vertices.

9. **Summarize the total distinct vertices involved:**
   - We have the initial vertex \( v \),
   - Plus \( k \) vertices in \( N(v) \),
   - Plus \( k(k-1) \) additional vertices from the neighbors of \( N(v) \) (excluding \( v \)).

10. **Total vertex count:**
    \[
    1 + k + k(k-1) = 1 + k + k^2 - k = 1 + k^2
    \]

11. **Final adjustment and conclusion:**
    - Note the minimum configuration must accommodate at least \( 2k \) vertices given that \( 1 + k^2 \geq 2k \) simplifies for \( k \geq 2 \), as \( k^2 \geq k \).

Thus, every \( k \)-regular graph with girth 4 must have at least \( 2k \) vertices.

Would you like more details or have any questions?

Here are five numbered questions you might want to ask next:
1. How can we construct a \( k \)-regular graph with girth 4?
2. What is an example of a 3-regular graph with girth 4?
3. Can a \( k \)-regular graph have a girth larger than 4?
4. What is the significance of the girth in graph theory?
5. How does the proof change for different values of \( k \)?

**Tip:** When studying graph theory, always pay attention to the definitions of graph properties like regularity and girth, as these form the foundation for many proofs and concepts.

Explore the proof demonstrating that every k-regular graph with girth 4 must have at least 2k vertices. This problem delves into graph theory concepts such as k-regular graphs and girth. Suitable for advanced high school students and beyond.

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