Given an integer $k$. $f(n)$ is defined on negative integer set and its values are integers. $f(n)$ satisfies \[ f(n)f(n+1)=(f(n)+n-k)^2, \] for $n=-2,-3,\cdots$. Find an expression of $f(n)$.
To find an expression for \( f(n) \) that satisfies the given functional equation:
\[ f(n)f(n+1) = (f(n) + n - k)^2 \]
for \( n = -2, -3, \ldots \), we will proceed step-by-step.

1. **Initial Assumption and Simplification:**
   Assume \( f(n) = (n - k)^2 \). We need to verify if this assumption satisfies the given functional equation.

2. **Substitute \( f(n) \) and \( f(n+1) \):**
   \[
   f(n) = (n - k)^2 \quad \text{and} \quad f(n+1) = (n + 1 - k)^2
   \]

3. **Substitute into the Functional Equation:**
   \[
   f(n)f(n+1) = (f(n) + n - k)^2
   \]
   Substituting the assumed forms:
   \[
   (n - k)^2 (n + 1 - k)^2 = ((n - k)^2 + n - k)^2
   \]

4. **Simplify the Right-Hand Side:**
   \[
   ((n - k)^2 + n - k)^2 = ((n - k)^2 + (n - k))^2
   \]
   \[
   = (n - k)^2 (n - k + 1)^2
   \]

5. **Verify the Equality:**
   \[
   (n - k)^2 (n + 1 - k)^2 = (n - k)^2 (n - k + 1)^2
   \]
   Since both sides are equal, our assumption \( f(n) = (n - k)^2 \) holds.

6. **Special Cases for \( k \):**
   - If \( k = -2 \), we have:
     \[
     f(n) = (n + 2)^2
     \]
     This works for all \( n \neq -1 \). For \( n = -1 \), we can have \( f(-1) = a \) where \( a \in \mathbb{Z} \).

   - If \( k = -3 \), we have:
     \[
     f(n) = (n + 3)^2
     \]
     This works for all \( n \neq -2 \). For \( n = -2 \), we can have \( f(-2) = -1 \) and \( f(-1) = 0 \).

7. **General Case:**
   For \( k \neq -2 \) and \( k \neq -3 \), the function \( f(n) = (n - k)^2 \) works for all \( n \).

Conclusion:
The functions that work are:
\[
\boxed{f(n) = (n - k)^2}
\]
If \( k = -2 \), we also have:
\[
\boxed{f(n) = (n - k)^2, k \neq -1, f(-1) = a, a \in \mathbb{Z}}
\]
If \( k = -3 \), we have:
\[
\boxed{f(n) = (n - k)^2, k > -2, f(-2) = -1, \text{and } f(-1) = 0}
\]