Determine all integers $a$ such that $a^k + 1$ is divisible by $12321$ for some $k$
To determine all integers \( a \) such that \( a^k + 1 \) is divisible by \( 12321 \) for some \( k \), we start by factoring \( 12321 \):

\[ 12321 = 3^2 \times 37^2 \]

We need \( a^k + 1 \equiv 0 \pmod{12321} \), which implies:

\[ a^k \equiv -1 \pmod{12321} \]

This can be broken down into two congruences:

\[ a^k \equiv -1 \pmod{9} \]
\[ a^k \equiv -1 \pmod{1369} \]

### Step 1: Solving \( a^k \equiv -1 \pmod{9} \)

Since \( 9 = 3^2 \), we consider the possible values of \( a \mod 3 \):

\[ a^2 \equiv 0, 1 \pmod{3} \]

For \( a^k \equiv -1 \pmod{9} \), \( k \) must be odd, and:

\[ a^k \equiv a \pmod{3} \]

Thus, \( a \equiv -1 \pmod{3} \). If \( a \equiv -1 \pmod{3} \), then:

\[ a^3 \equiv -1 \pmod{9} \]

### Step 2: Solving \( a^k \equiv -1 \pmod{1369} \)

Since \( 1369 = 37^2 \), we need:

\[ a^k \equiv -1 \pmod{37^2} \]

First, consider \( a^k \equiv -1 \pmod{37} \). By Fermat's Little Theorem:

\[ a^{36} \equiv 1 \pmod{37} \]

Suppose the smallest positive integer \( d \) such that \( a^d \equiv 1 \pmod{37} \). Then \( d \) must divide \( 2k \) and \( 36 \), and \( d \) must be even. Let \( d = 2t \) where \( t \) is odd. Then:

\[ a^t \equiv -1 \pmod{37} \]

Since \( t \) must divide 18 (the half of 36), \( t \) can be 1, 3, 9. Therefore:

\[ a^9 \equiv -1 \pmod{37} \]

### Step 3: Combining the Conditions

We need \( a^9 \equiv -1 \pmod{37} \) and \( a \equiv -1 \pmod{3} \). The solutions to \( a^9 \equiv -1 \pmod{37} \) are:

\[ a \equiv -1, 3, 4, -7, -9, -10, 11, -12, -16 \pmod{37} \]

### Step 4: Finding Common Solutions

We need to find \( a \) such that:

\[ a \equiv -1 \pmod{3} \]

Checking each solution modulo 37:

- \( -1 \equiv 2 \pmod{3} \)
- \( 3 \equiv 0 \pmod{3} \)
- \( 4 \equiv 1 \pmod{3} \)
- \( -7 \equiv 2 \pmod{3} \)
- \( -9 \equiv 0 \pmod{3} \)
- \( -10 \equiv 2 \pmod{3} \)
- \( 11 \equiv 2 \pmod{3} \)
- \( -12 \equiv 0 \pmod{3} \)
- \( -16 \equiv 2 \pmod{3} \)

The valid solutions are:

\[ a \equiv 11, 41, 62, 65, 77, 95, 101, 104, 110 \pmod{111} \]

The final answer is \( \boxed{ a \equiv 11, 41, 62, 65, 77, 95, 101, 104, 110 \pmod{111} } \).