Solution:
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.

Solution:
Let $p$ be a prime such that $p$ divides $b$. For some integer $x$, we have
$$(x-1)^{5} \equiv x^{5} \equiv(x+1)^{5} \quad(\bmod p)$$
Now, there is a primitive root $g$ modulo $p$, so there exist $u, v, w$ such that
$$x-1 \equiv g^{u} \quad(\bmod p), \quad x \equiv g^{v} \quad(\bmod p), \quad x+1 \equiv g^{w} \quad(\bmod p)$$
The condition of the problem is thus
$$g^{5 u} \equiv g^{5 v} \equiv g^{5 w} \quad(\bmod p) \quad \Longrightarrow \quad 5 u \equiv 5 v \equiv 5 w \quad(\bmod p-1)$$
If $p \not \equiv 1(\bmod 5)$, then 5 is invertible modulo $p-1$ and thus $u \equiv v \equiv w(\bmod p-1)$, i.e. $x-1 \equiv x \equiv x+1(\bmod p)$. This is a contradiction. Hence $p \equiv 1(\bmod 5)$ and thus $u \equiv v \equiv w\left(\bmod \frac{p-1}{5}\right)$. Thus, from (6), there exist integers $k, \ell$ such that
$$\left.\begin{array}{rl} x-1 & \equiv g^{v+k \frac{p-1}{5}} \equiv x t^{k} \quad(\bmod p) \\ x+1 & \equiv g^{v+\ell \frac{p-1}{5}} \\ \equiv x t^{\ell} & (\bmod p) \end{array}\right\} \quad \text { where } t=g^{\frac{p-1}{5}}$$
Let $r=t^{k}$ and $s=t^{\ell}$. In particular, the above yields $r, s \not \equiv 1(\bmod p)$, and thus
$$x \equiv-(r-1)^{-1} \equiv(s-1)^{-1} \quad(\bmod p)$$
It follows that
$$(r-1)^{-1}+(s-1)^{-1} \equiv 0 \quad(\bmod p) \quad \Longrightarrow \quad r+s \equiv 2 \quad(\bmod p)$$
Now $t^{5} \equiv 1(\bmod p)$, so $r$ and $s$ must be congruent, modulo $p$, to some of the non-trivial fifth roots of unity $t, t^{2}, t^{3}, t^{4}$. Observe that, for any pair of these non-trivial roots of unity,
either one is the other's inverse, or one is the other's square. In the first case, we have $r+r^{-1} \equiv 2(\bmod p)$, implying $r \equiv 1(\bmod p)$, a contradiction. Hence
$$r+r^{2} \equiv 2 \quad(\bmod p) \quad \Longrightarrow \quad(r-1)(r+2) \equiv 0 \quad(\bmod p),$$
or
$$s^{2}+s \equiv 2 \quad(\bmod p) \quad \Longrightarrow \quad(s-1)(s+2) \equiv 0 \quad(\bmod p)$$
Thus, since $r, s \not \equiv 1(\bmod p)$, we have $r \equiv-2(\bmod p)$ or $s \equiv-2(\bmod p)$, and thus $1 \equiv r^{5} \equiv-32(\bmod p)$ or an analogous equation obtained from $s$. Hence $p \mid 33$. Since $p \equiv 1(\bmod 5)$, it follows that $p=11$, i.e. $b$ is a power of 11 .
Examining the fifth powers modulo 11 , we see that $b=11$ is indeed a solution with $a \equiv \pm 1(\bmod 11)$ and, correspondingly, $x \equiv \pm 4(\bmod 11)$. Now suppose, for the sake of contradiction, that $11^{2}$ divides $b$. Then, for some integer $m$, we must have
$$(x-1, x, x+1) \equiv \pm(3+11 m, 4+11 m, 5+11 m) \quad(\bmod 121)$$
and thus, substituting into the condition of the problem,
$$\begin{aligned} 3^{5}+55 \cdot 3^{4} m & \equiv 4^{5}+55 \cdot 4^{4} m \equiv 5^{5}+55 \cdot 5^{4} m \quad(\bmod 121) \\ & \Longrightarrow \quad 1-22 m \equiv 56+44 m \equiv-21+11 m \quad(\bmod 121) \end{aligned}$$
Hence $33 m \equiv 22(\bmod 121)$ and $33 m \equiv 44(\bmod 121)$, so $22 \equiv 0(\bmod 121)$, a contradiction. It follows that $b \mid 11$.
Finally, we conclude that the positive integers satisfying the original condition are $b=11$, with $a \equiv \pm 1(\bmod 11)$, and $b=1$, for any positive integer $a$.

Solution:
Denote the three consecutive integers by $x-1, x$, and $x+1$ as in Solution 1. By computing the differences in (1), we find
$$\begin{aligned} & F:=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \equiv 0 \quad(\bmod b), \\ & G:=x^{5}-(x-1)^{5}=5 x^{4}-10 x^{3}+10 x^{2}-5 x+1 \equiv 0 \quad(\bmod b) . \end{aligned}$$
By determining the polynomial greatest divisor of $F(x)$ and $G(x)$ using the Euclidean algorithm, we find that
$$p(x) F(x)+q(x) G(x)=22$$
where
$$\begin{aligned} & p(x)=-15 x^{3}+30 x^{2}-28 x+11 \\ & q(x)=15 x^{3}+30 x^{2}+28 x+11 \end{aligned}$$
Since $b \mid F(x)$ and $b \mid G(x)$, it follows from (7) that $b \mid 22$. We now finish off the problem as in Solution 1.