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.

1. Define the polynomial and set up the problem:
We are given the polynomial $P(n) = \frac{n^5 + a}{b}$ and need to find all positive integers $a$ and $b$ such that $P(n)$ takes integer values for three consecutive integers $n$.

2. **Introduce a new polynomial $Q(n)$:**
Let $Q(n) = n^5 + a$. For $P(n)$ to be an integer, $b$ must divide $Q(n)$ for three consecutive integers $n$.

3. Analyze the divisibility conditions:
We need $b$ to divide the differences between $Q(n)$ evaluated at consecutive integers:
$$b \mid Q(n+1) + Q(n-1) - 2Q(n)$$
$$b \mid Q(n+1) - Q(n-1)$$

4. Calculate the differences:
$$Q(n+1) = (n+1)^5 + a = n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 + a$$
$$Q(n-1) = (n-1)^5 + a = n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1 + a$$
$$Q(n+1) + Q(n-1) - 2Q(n) = (n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 + a) + (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1 + a) - 2(n^5 + a)$$
Simplifying, we get:
$$Q(n+1) + Q(n-1) - 2Q(n) = 20n^3 + 10n$$
Therefore:
$$b \mid 20n^3 + 10n$$

5. Calculate the second difference:
$$Q(n+1) - Q(n-1) = (n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 + a) - (n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1 + a)$$
Simplifying, we get:
$$Q(n+1) - Q(n-1) = 10n^4 + 20n^2 + 2$$
Therefore:
$$b \mid 10n^4 + 20n^2 + 2$$

6. Combine the conditions:
$$b \mid 2(10n^4 + 20n^2 + 2) - n(20n^3 + 10n)$$
Simplifying, we get:
$$b \mid 20n^4 + 40n^2 + 4 - 20n^4 - 10n^2 = 30n^2 + 4$$
Therefore:
$$b \mid 30n^2 + 4$$

7. Further simplification:
$$b \mid 2n(30n^2 + 4) - 3(20n^3 + 10n)$$
Simplifying, we get:
$$b \mid 60n^3 + 8n - 60n^3 - 30n = -22n$$
Therefore:
$$b \mid 22n$$

8. **Determine possible values of $b$:**
Since $b$ must divide $22n$ for all $n$, and $\gcd(Q(n+1) - Q(n-1), n) = 1$, we have $\gcd(b, n) = 1$. Thus, $b$ must divide 22. The possible values for $b$ are 1, 2, 11, and 22.

9. Case analysis:
- **Case 1: $b = 1$**
Any integer value of $a$ works since $P(n) = n^5 + a$ is always an integer.
- **Case 2: $b = 2$
We cannot have three consecutive fifth powers with the same remainder modulo 2, so no solution exists.
- Case 3: $b = 11$**
We need $n^5 + a \equiv 0 \pmod{11}$ for three consecutive $n$. Observing the fifth powers modulo 11:
$$1^5 \equiv 1, \quad 2^5 \equiv -1, \quad 3^5 \equiv 1, \quad 4^5 \equiv -1, \quad 5^5 \equiv 1 \pmod{11}$$
We find that $a \equiv -1 \pmod{11}$ or $a \equiv 1 \pmod{11}$.
- **Case 4: $b = 22$
By the Chinese Remainder Theorem, we cannot have three consecutive fifth powers with the same remainder modulo 22.

10. Conclusion:**
The only solutions are $(a, b) = (k, 1)$ and $(a, b) = (11k \pm 1, 11)$ for $k \in \mathbb{Z}$.

The final answer is $\boxed{ (a, b) = (k, 1), (11k \pm 1, 11) }$ for $k \in \mathbb{Z}$.