Let $k$ be a given integer, and $f(n)$ be a function defined on the set of negative integers with integer values, satisfying
$$
\begin{array}{l}
f(n) f(n+1)=(f(n)+n-k)^{2}, \\
n=-2,-3,-4, \cdots .
\end{array}
$$

Find the expression for the function $f(n)$.
Lemma: There exist infinitely many prime numbers that are not of the form $5k \pm 1$.

Proof: For any given positive integer $n$, consider
$N=5[1 \cdot 3 \cdot 5 \cdots \cdots(2n+1)]^{2}+2$.
Since the product of integers of the form $5k \pm 1$ is still of the form $5k \pm 1$, $N \equiv 2(\bmod 5)$, hence $N$ has a prime factor of the form $5k \pm 2$, which is greater than $2n+1$. The lemma is proved.

Take a prime $p>10(|k|+1), p \neq \pm 1(\bmod 5)$. First, we prove $f(k-p)=p^{2}$.
From $f(k-p) f(k-p+1)=[f(k-p)-p]^{2}$, we know $f(k-p) \mid p^{2}$.
Therefore, $f(k-p)= \pm 1, \pm p, \pm p^{2}$.
Notice that
$f(k-p-1) f(k-p)=[f(k-p-1)-p-1]^{2}$.
Considering $a x=(x-p-1)^{2}$,

i.e., $x^{2}-(2p+2+a) x+(p+1)^{2}=0$.
Its discriminant is
$\Delta(a)=(2p+2+a)^{2}-4(p+1)^{2}=a(a+4p+4)$.
Since $\Delta(1)=4p+5 \equiv \pm 2(\bmod 5)$,
$\Delta(-1)=-(4p+3)<0$,
$\Delta(p)=p(5p+4)$ is not a perfect square.
$\Delta(-p)=-p(3p+4)<0$.
$\Delta\left(-p^{2}\right)=p^{2}\left(p^{2}-4p-4\right) \equiv \pm 2(\bmod 5)$.
Thus, $\Delta(1), \Delta(-1), \Delta(p), \Delta(-p), \Delta\left(-p^{2}\right)$

are all not perfect squares. Hence,
$f(k-p)=p^{2}$.
From $f(k-p) f(k-p+1)=[f(k-p)-p]^{2}$, we know
$f(k-p+1)=(p-1)^{2}$.
From $f(k-p+1) f(k-p+2)=[f(k-p+1)-p+1]^{2}$, we know
$f(k-p+2)=(p-2)^{2}$.
Continuing in this manner, we get
$f(k-p+t)=(p-t)^{2}, 0 \leqslant t \leqslant p$ and $t<p-k$,

i.e., $f(n)=(n-k)^{2}, k-p \leqslant n \leqslant k$ and $n<0$.
Since $p$ can be arbitrarily large, we have
$f(n)=(n-k)^{2}, n \leqslant k$ and $n<0$.
We now discuss based on the value of $k$
Case 1: $k \geqslant-1$. From the above conclusion, we know
$f(n)=(n-k)^{2}, n=-1,-2, \cdots$.
Case 2: $k=-2$. From the above conclusion, we know
$f(n)=(n-k)^{2}, n=-2,-3, \cdots$.
From $f(-2) f(-1)=[f(-2)+0]^{2}$, we know $f(-1)$ can be any integer. In this case,
$$
f(n)=\left\{\begin{array}{ll}
a, & n=-1, \\
(n+2)^{2}, & n=-2,-3, \cdots .
\end{array}\right.
$$

where $a$ is any integer.
Case 3: $k=-3$. In this case,
$$
\begin{array}{l}
f(n)=(n-k)^{2}, n=-3,-4, \cdots . \\
\text { Also, } f(-3) f(-2)=[f(-3)+0]^{2}, \\
f(-2) f(-1)=[f(-2)+1]^{2},
\end{array}
$$

so $f(-2)= \pm 1$.
If $f(-2)=1$, then $f(-1)=2^{2}$;
If $f(-2)=-1$, then $f(-1)=0$.
In this case, $f(n)=\left\{\begin{array}{ll}0, & n=-1, \\ -1, & n=-2 \\ (n+3)^{2}, & n \leqslant-3\end{array}\right.$
or $f(n)=\left\{\begin{array}{ll}2^{2}, & n=-1, \\ 1, & n=-2, \\ (n+3)^{2}, & n \leqslant-3 .\end{array}\right.$
Case 4: $k \leqslant-4$.
From $f(k+1) f(k+2)=[f(k+1)+1]^{2}$

we know $f(k+1)= \pm 1$.
Assume $f(k+1)=-1$, from (1) we know $f(k+2)=0$.
Also, $f(k+2) f(k+3)=[f(k+2)+2]^{2}$,
then $0=2^{2}$, which is a contradiction. Therefore, $f(k+1)=1$.
From (1) we know $f(k+2)=2^{2}$,
From $f(k+2) f(k+3)=[f(k+2)+2]^{2}$ we know
$f(k+3)=3^{2}$.

Continuing in this manner, we have
$$
f(k+t)=t^{2}, t \leqslant-k-1 .
$$

In this case, $f(n)=(n-k)^{2}, n=-1,-2, \cdots$