Problem 2288

Let us consider the problem of finding all positive integers $n$ for which there exists a unique integer $a$ such that $0 \leq a < n!$ and

$$n! \mid a^n + 1.$$

### Step-by-step Solution:

1. Understand the Divisibility Condition:

We require that $a^n + 1 \equiv 0 \pmod{n!}$, meaning:

$$a^n \equiv -1 \pmod{n!}.$$

2. Explore Special Cases and General Patterns:

**Case $n = 1$:**

- For $n = 1$, we seek $0 \leq a < 1! = 1$, so $a = 0$.
- Then, $a^1 + 1 = 0^1 + 1 = 1 \equiv 0 \pmod{1}$, which holds.

Hence, $n = 1$ is a solution.

**Case $n$ is a Prime:**

- Let $n$ be a prime number.
- Wilson's Theorem states $(n-1)! \equiv -1 \pmod{n}$, implying for $a = n-1$, we have:
$$(n-1)^n = (n-1)^{n-1} \cdot (n-1) \equiv (-1)^{n-1} \cdot (n-1) \equiv -1 \equiv 0 \pmod{n}.$$
- We need $(n-1)^n + 1 \equiv 0 \pmod{n!}$.
- Notice if $k = n-1$, $(n-1)! \equiv -1 \pmod{n}$ implies:

$$(n-1)^n \equiv -1 \equiv 0 \pmod{n!}.$$

- Unique $a = n-1$ exists and satisfies the conditions for primes.

Thus, all prime numbers $n$ also satisfy the condition as they create a unique choice for $a = n-1$.

3. Check if Further Conditions Can Be Satisfied:

- For composite $n$, any $a$ less than $n!$ that works has non-uniqueness due to additional factors canceling divisors.

4. Conclusion:

By examining divisibility and uniqueness conditions, we find that:

$$\boxed{\text{All prime numbers or } n = 1}$$

These are the solutions where a unique $a$ can be found satisfying the given condition.