Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial

$$
P(n)=\frac{n^{5}+a}{b}
$$

takes integer values.
Denote the three consecutive integers by \(x-1, x\), and \(x+1\), so that

\[
(x-1)^{5}+a \equiv 0 \quad(\bmod b), \quad x^{5}+a \equiv 0 \quad(\bmod b), \quad(x+1)^{5}+a \equiv 0 \quad(\bmod b) .
\]

By computing the differences of the equations in (1) we get

\[
\begin{aligned}
A & :=(x+1)^{5}-(x-1)^{5}=10 x^{4}+20 x^{2}+2 \equiv 0 \quad(\bmod b) \\
B & :=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \equiv 0 \quad(\bmod b)
\end{aligned}
\]

Adding the first and third equation in (1) and subtracting twice the second equation yields

\[
C:=(x+1)^{5}+(x-1)^{5}-2 x^{5}=20 x^{3}+10 x \equiv 0 \quad(\bmod b) .
\]

Next, (2) and (4) together yield

\[
D:=4 x A-\left(2 x^{2}+3\right) C=-22 x \equiv 0(\bmod b) .
\]

Finally we combine (3) and (5) to derive

\[
22 B+\left(5 x^{3}+10 x^{2}+10 x+5\right) D=22 \equiv 0(\bmod b) .
\]

As the positive integer \(b\) divides 22, we are left with the four cases \(b=1, b=2, b=11\) and \(b=22\).

If \(b\) is even (i.e., \(b=2\) or \(b=22\)), then we get a contradiction from (3), because the integer \(B=2\left(5 x^{3}+5 x^{2}\right)+5\left(x^{4}+x\right)+1\) is odd, and hence not divisible by any even integer.

For \(b=1\), it is trivial to see that a polynomial of the form \(P(n)=n^{5}+a\), with \(a\) any positive integer, has the desired property.

For \(b=11\), we note that

\[
\begin{aligned}
& n \equiv 0,1,2,3,4,5,6,7,8,9,10 \quad(\bmod 11) \\
& \Longrightarrow \quad n^{5} \equiv 0,1,-1,1,1,1,-1,-1,-1,1,-1 \quad(\bmod 11)
\end{aligned}
\]

Hence a polynomial of the form \(P(n)=\left(n^{5}+a\right) / 11\) has the desired property if and only if \(a \equiv \pm 1(\bmod 11)\). This completes the proof.

A Variant. We start by following the first solution up to equation (4). We note that \(b=1\) is a trivial solution, and assume from now on that \(b \geqslant 2\). As \((x-1)^{5}+a\) and \(x^{5}+a\) have different parity, \(b\) must be odd. As \(B\) in (3) is a multiple of \(b\), we conclude that (i) \(b\) is not divisible by 5 and that (ii) \(b\) and \(x\) are relatively prime. As \(C=10 x\left(2 x^{2}+1\right)\) in (4) is divisible by \(b\), we altogether derive

\[
E:=2 x^{2}+1 \equiv 0(\bmod b)
\]

Together with (2) this implies that

\[
5 E^{2}+10 E-2 A=11 \equiv 0 \quad(\bmod b)
\]

Hence \(b=11\) is the only remaining candidate, and it is handled as in the first solution.