Problem: Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial $P(n) = \frac1b(n^5+a)$ takes integer values.

Solution: The answer is
\begin{itemize}
  \ii Either $b = 1$; or
  \ii $a \equiv \pm 1 \pmod{11}$ and $b = 11$ (and $a > 0$).
\end{itemize}

We consider only the case $b>1$.
Suppose that $P$ is a prime power dividing $b$.
Evidently $P$ is not even.
Then there exists an $n$ for which
\[ n^5 \equiv (n+1)^5 \equiv (n-1)^5 \pmod{P}.  \]
Therefore,
\[ 0 \equiv (n+1)^5 + (n-1)^5 - 2n^5 = 20n^3+10n \pmod{P}.  \]
Since $n \not\equiv 0 \pmod{p}$ for obvious reasons, we obtain that
\[ n^2 \equiv -\tfrac{1}{2} \pmod{p}. \]
Then,
\begin{align*}
  0 &\equiv 2((n+1)^5 - (n-1)^5) \pmod{P} \\
  &= 20n^4+40n^2+4 \\
  &\equiv 20 \cdot -\frac{1}{4} + 40 \cdot -\frac{1}{2} + 4 \pmod{P} \\
  &= -11.
\end{align*}
This implies $P = 11$.
Since $P$ can be any prime power dividing $b$, this forces $b=11$.

In other words, $b \in \left\{ 1,11 \right\}$.

Finally, observe that
\[ 3^5 \equiv 4^5 \equiv 5^5 \equiv 1 \pmod{11} \]
and \[ (-3)^5 \equiv (-4)^5 \equiv (-5)^5 \equiv -1 \pmod{11} \]
are the only such triples modulo $11$.
Therefore the solution set is the one claimed above.