Find all positive integers $a, b$, and $c$ such that the numbers
$$\frac{a+1}{b}, \frac{b+1}{c} \quad \text{and} \quad \frac{c+1}{a}$$
are positive integers.
To find all positive integers \(a, b, c\) such that the numbers \(\frac{a+1}{b}, \frac{b+1}{c}, \frac{c+1}{a}\) are positive integers, we start by noting the conditions:

1. \(b \mid (a+1)\)
2. \(c \mid (b+1)\)
3. \(a \mid (c+1)\)

From these conditions, we can derive the following inequalities:
\[ a \leq c+1 \]
\[ c \leq b+1 \]
\[ b \leq a+1 \]

Combining these, we get:
\[ a \leq c+1 \leq b+2 \leq a+3 \]

We will consider the cases \(a \leq b \leq c\) and \(a \leq c \leq b\) separately.

### Case I: \(a \leq b \leq c\)
1. **Subcase \(b = c\):**
   - If \(b = c\), then \(b \mid (a+1)\) and \(a \mid (b+1)\).
   - Since \(b \leq a+1\) and \(a \leq b+1\), we have \(a \leq b \leq a+1\).
   - This implies \(b = a\) or \(b = a+1\).

   - If \(b = a\), then \(a \mid (a+1)\), which is only true if \(a = 1\). Thus, \(a = b = c = 1\).

   - If \(b = a+1\), then \(a \mid (a+2)\), which is true for \(a = 1\) or \(a = 2\). For \(a = 1\), we get \(b = 2\) and \(c = 2\), but this does not satisfy \(c \mid (b+1)\). For \(a = 2\), we get \(b = 3\) and \(c = 3\), but this does not satisfy \(c \mid (b+1)\).

2. **Subcase \(c = b+1\):**
   - If \(c = b+1\), then \(b \mid (a+1)\) and \(a \mid (b+2)\).
   - Since \(a \leq b \leq a+1\), we have \(a = b\) or \(a = b-1\).

   - If \(a = b\), then \(b \mid (b+1)\), which is not possible.

   - If \(a = b-1\), then \(b-1 \mid (b+2)\), which implies \(b-1 \mid 3\). Thus, \(b-1 = 1\) or \(b-1 = 3\). This gives \(b = 2\) or \(b = 4\).

     - For \(b = 2\), we get \(a = 1\) and \(c = 3\), but this does not satisfy \(c \mid (b+1)\).

     - For \(b = 4\), we get \(a = 3\) and \(c = 5\). This satisfies all conditions: \(4 \mid (3+1)\), \(5 \mid (4+1)\), and \(3 \mid (5+1)\).

### Case II: \(a \leq c \leq b\)
1. **Subcase \(b = a+1\):**
   - If \(b = a+1\), then \(c \mid (a+2)\) and \(a \mid (c+1)\).
   - Since \(a \leq c \leq a+1\), we have \(c = a\) or \(c = a+1\).

   - If \(c = a\), then \(a \mid (a+1)\), which is only true if \(a = 1\). Thus, \(a = c = 1\) and \(b = 2\).

   - If \(c = a+1\), then \(a \mid (a+2)\), which is true for \(a = 1\) or \(a = 2\). For \(a = 1\), we get \(b = 2\) and \(c = 2\), but this does not satisfy \(c \mid (b+1)\). For \(a = 2\), we get \(b = 3\) and \(c = 3\), but this does not satisfy \(c \mid (b+1)\).

2. **Subcase \(c = a\):**
   - If \(c = a\), then \(a \mid (a+1)\), which is only true if \(a = 1\). Thus, \(a = b = c = 1\).

### Conclusion:
The only solutions are \((1, 1, 1)\) and \((3, 4, 5)\) along with their cyclic permutations.

The final answer is \( \boxed{ (1, 1, 1) } \), \((3, 4, 5)\), and their cyclic permutations.