Four, (50 points) Find all positive integers $a, b$ such that the polynomial
$$
f(x)=\frac{x^{5}+a}{b}
$$
Four, $f(n) \in \mathbf{Z}$ if and only if $b \mid\left(n^{5}+a\right)$.
Obviously, $b=1$ (for any $a$) satisfies the condition.
Assume $b>1$.
Suppose $f(n-1) 、 f(n) 、 f(n+1) \in \mathbf{Z}$.
Then $(n-1)^{5}+a 、 n^{5}+a 、(n+1)^{5}+a$ are all divisible by $b$.
These three numbers alternate in parity, hence $b$ is odd.
Also, $b \mid\left[(n+1)^{5}-n^{5}\right]$, i.e.,
$$
b \mid\left[5 n^{4}+10 n^{3}+10 n^{2}+5 n+1\right] \text {, }
$$

Then $b$ and $5 n$ are coprime.
$$
\begin{array}{l}
\text { Additionally, } b \mid\left[(n-1)^{5}+(n+1)^{5}-2 n^{5}\right. \\
\Rightarrow b\left|\left(20 n^{3}+10 n\right) \Rightarrow b\right| 10 n\left(2 n^{2}+1\right) .
\end{array}
$$

But $b$ and $10 n$ are coprime, so $b \mid\left(2 n^{2}+1\right)$, i.e.,
$$
\begin{array}{l}
2 n^{2} \equiv-1(\bmod b) . \\
\text { Then } 0 \equiv 4\left[(n+1)^{5}-n^{5}\right] \\
=5\left(2 n^{2}\right)^{2}+20 n\left(2 n^{2}\right)+20\left(2 n^{2}\right)+20 n+4 \\
\equiv-11(\bmod b) \\
\Rightarrow b \mid 11 .
\end{array}
$$

Therefore, the only solution greater than 1 is $b=11$.
When $b=11$, for integers $x$ coprime with 11,
$$
x^{10} \equiv 1(\bmod 11) \text {, }
$$

Thus, $x^{5} \equiv \pm 1(\bmod 11)$.
Therefore, $a \equiv \pm 1(\bmod 11)$.
And by $3^{5} \equiv 4^{5} \equiv 5^{5} \equiv 1(\bmod 11)$,
$$
6^{5} \equiv 7^{5} \equiv 8^{5} \equiv-1(\bmod 11),
$$

When $a \equiv 1(\bmod 11)$, take $n=7$, and when $a \equiv-1(\bmod 11)$, take $n=4$ to satisfy the condition.

Hence, all solutions are $b=1$, $a$ being any integer, or $b=11$, $a$ being any positive integer that is congruent to $\pm 1$ modulo $b$.