Find all possible triples of positive integers, $a, b, c$ so that $\frac{a+1}{b}$, $\frac{b+1}{c}$ and $\frac{c+1}{a}$ are also integers.
To find all possible triples of positive integers \( (a, b, c) \) such that \( \frac{a+1}{b} \), \( \frac{b+1}{c} \), and \( \frac{c+1}{a} \) are integers, we start by analyzing the given conditions.

1. **Initial Conditions:**
   \[
   \frac{a+1}{b} \in \mathbb{Z}, \quad \frac{b+1}{c} \in \mathbb{Z}, \quad \frac{c+1}{a} \in \mathbb{Z}
   \]
   This implies:
   \[
   b \mid a+1, \quad c \mid b+1, \quad a \mid c+1
   \]

2. **Inequalities:**
   From the divisibility conditions, we get:
   \[
   a \leq c+1, \quad b \leq a+1, \quad c \leq b+1
   \]

3. **Assumption and Contradiction:**
   Assume \( a < c \), \( b < a \), and \( c < b \). Adding these inequalities:
   \[
   a < c, \quad b < a, \quad c < b \implies a < b < c < a
   \]
   This leads to \( a < a \), which is a contradiction. Therefore, at least one of the inequalities must be an equality.

4. **Case Analysis:**
   We analyze the cases where at least one of the variables is equal to another or differs by 1.

   - **Case 1: \( a = c \) or \( a = c + 1 \)**
     - **Subcase 1.1: \( a = c \)**
       \[
       \frac{a+1}{b}, \quad \frac{b+1}{a}, \quad \frac{a+1}{a} \text{ must be integers}
       \]
       The third fraction \( \frac{a+1}{a} \) is an integer if and only if \( a = 1 \). Thus, \( c = 1 \) and:
       \[
       \frac{2}{b} \text{ must be an integer} \implies b = 1 \text{ or } 2
       \]
       Therefore, the solutions are \( (1, 1, 1) \) and \( (1, 2, 1) \).

     - **Subcase 1.2: \( a = c + 1 \)**
       \[
       \frac{c+2}{b}, \quad \frac{b+1}{c} \text{ must be integers}
       \]
       This implies:
       \[
       b \mid c+2 \quad \text{and} \quad c \mid b+1
       \]
       We need to find values of \( b \) and \( c \) that satisfy these conditions.

   - **Case 2: \( b = a \) or \( b = a + 1 \)**
     - **Subcase 2.1: \( b = a \)**
       \[
       \frac{a+1}{a}, \quad \frac{a+1}{c}, \quad \frac{c+1}{a} \text{ must be integers}
       \]
       The first fraction \( \frac{a+1}{a} \) is an integer if and only if \( a = 1 \). Thus, \( b = 1 \) and:
       \[
       \frac{2}{c} \text{ must be an integer} \implies c = 1 \text{ or } 2
       \]
       Therefore, the solutions are \( (1, 1, 1) \) and \( (1, 1, 2) \).

     - **Subcase 2.2: \( b = a + 1 \)**
       \[
       \frac{a+2}{c}, \quad \frac{c+1}{a} \text{ must be integers}
       \]
       This implies:
       \[
       c \mid a+2 \quad \text{and} \quad a \mid c+1
       \]
       We need to find values of \( a \) and \( c \) that satisfy these conditions.

   - **Case 3: \( c = b \) or \( c = b + 1 \)**
     - **Subcase 3.1: \( c = b \)**
       \[
       \frac{a+1}{b}, \quad \frac{b+1}{b}, \quad \frac{b+1}{a} \text{ must be integers}
       \]
       The second fraction \( \frac{b+1}{b} \) is an integer if and only if \( b = 1 \). Thus, \( c = 1 \) and:
       \[
       \frac{2}{a} \text{ must be an integer} \implies a = 1 \text{ or } 2
       \]
       Therefore, the solutions are \( (1, 1, 1) \) and \( (2, 1, 1) \).

     - **Subcase 3.2: \( c = b + 1 \)**
       \[
       \frac{a+1}{b}, \quad \frac{b+2}{a} \text{ must be integers}
       \]
       This implies:
       \[
       a \mid b+2 \quad \text{and} \quad b \mid a+1
       \]
       We need to find values of \( a \) and \( b \) that satisfy these conditions.

5. **Conclusion:**
   After analyzing all cases, the valid solutions are:
   \[
   (1, 1, 1), \quad (1, 2, 1), \quad (1, 1, 2), \quad (2, 1, 1)
   \]

The final answer is \( \boxed{(1, 1, 1), (1, 2, 1), (1, 1, 2), (2, 1, 1)} \).