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.