Solution:
Let $f(x)$ be as given, and suppose $f(a) = f(b) = f(c) = f(d) = 5$ for four distinct integers $a, b, c, d$.

Consider the polynomial $g(x) = f(x) - 5$. Then $g(a) = g(b) = g(c) = g(d) = 0$, so $a, b, c, d$ are roots of $g(x)$.

Therefore, $g(x)$ is divisible by $(x - a)(x - b)(x - c)(x - d)$, i.e.,
$$g(x) = (x - a)(x - b)(x - c)(x - d) h(x)$$
for some polynomial $h(x)$ with integer coefficients (since $f(x)$ has integer coefficients and $5$ is an integer).

Suppose there exists an integer $k$ such that $f(k) = 8$. Then $g(k) = f(k) - 5 = 3$.

So,
$$g(k) = (k - a)(k - b)(k - c)(k - d) h(k) = 3$$
But $k, a, b, c, d$ are all integers, so $(k - a)(k - b)(k - c)(k - d)$ is an integer, and $h(k)$ is an integer.

Therefore, $(k - a)(k - b)(k - c)(k - d) h(k) = 3$.

But $3$ has only the following possible factorizations into integer factors:

- $3 = 1 \times 1 \times 1 \times 3$
- $3 = 1 \times 1 \times (-1) \times (-3)$
- $3 = 1 \times (-1) \times 1 \times (-3)$
- $3 = (-1) \times 1 \times 1 \times (-3)$
- $3 = (-1) \times (-1) \times (-1) \times (-3)$
- $3 = 3 \times 1 \times 1 \times 1$
- $3 = (-3) \times 1 \times 1 \times (-1)$
- $3 = 1 \times 3 \times 1 \times 1$
- $3 = 1 \times 1 \times 3 \times 1$
- $3 = 1 \times 1 \times 1 \times 3$

But in all cases, since $a, b, c, d$ are distinct integers, $k - a$, $k - b$, $k - c$, $k - d$ are four distinct nonzero integers. The product of four distinct nonzero integers cannot be $\pm 1$ or $\pm 3$.

To see this, note that the absolute value of the product $| (k - a)(k - b)(k - c)(k - d) |$ is at least $1 \times 2 \times 3 \times 4 = 24$ (if the differences are $\pm 1, \pm 2, \pm 3, \pm 4$), which is much greater than $3$.

Therefore, there is no integer $k$ such that $f(k) = 8$.