Let $p$ be a prime number and $F=\left \{0,1,2,...,p-1 \right \}$. Let $A$ be a proper subset of $F$ that satisfies the following property:  if $a,b \in A$, then $ab+1$ (mod $p$) $ \in A$.
How many elements can $A$ have? (Justify your answer.)

1. **Initial Considerations:**
   - Let \( p \) be a prime number and \( F = \{0, 1, 2, \ldots, p-1\} \).
   - Let \( A \) be a proper subset of \( F \) such that if \( a, b \in A \), then \( ab + 1 \pmod{p} \in A \).
   - We need to determine the possible sizes of \( A \).

2. **Case \( p = 2 \):**
   - If \( p = 2 \), then \( F = \{0, 1\} \).
   - Since \( A \) is a proper subset of \( F \), \( A \) can be either \( \{0\} \) or \( \{1\} \).
   - However, neither \( \{0\} \) nor \( \{1\} \) satisfies the condition \( ab + 1 \pmod{2} \in A \) for \( a, b \in A \).
   - Therefore, \( |A| = 0 \) for \( p = 2 \).

3. **Case \( p > 2 \):**
   - Assume \( p > 2 \) and \( |A| \geq 2 \).
   - Note that \( 0, 1 \notin A \) because if \( 0 \in A \), then \( 0 \cdot a + 1 = 1 \in A \), which contradicts \( 1 \notin A \).

4. **Cycle Decomposition:**
   - Consider the map \( m_{a,b}: x \mapsto ax + b \) where \( a, b \in F \) and \( a \neq 0 \).
   - For \( b = 1 \) and \( a \in A \), \( m_{a,1}(x) = ax + 1 \).
   - We need to analyze the cycle structure of \( m_{a,1} \).

5. **Multiplicative Order:**
   - Let \( \delta(a) \) be the multiplicative order of \( a \) modulo \( p \).
   - The map \( m_{a,1} \) has cycles of length \( 1 \) or \( \delta(a) \).
   - For \( m_{a,1} \) to map \( A \) onto itself, \( \delta(a) \) must divide the size of \( A \).

6. **Contradiction:**
   - Let \( L = \text{lcm}(\delta(a) \mid a \in A) \).
   - If \( |A| \geq 2 \), then \( |A| \geq L \).
   - The number of elements in \( F \) with order dividing \( L \) is at most \( L - 1 \).
   - This implies \( |A| \leq L - 1 \), which contradicts \( |A| \geq L \).

7. **Conclusion:**
   - The only possible size for \( A \) is \( |A| = 1 \) when \( p > 2 \).
   - For \( p = 3 \), \( A \) can be \( \{2\} \) since \( 2 \cdot 2 + 1 \equiv 2 \pmod{3} \).

The final answer is \( \boxed{ |A| = 1 } \) for \( p > 2 \) and \( |A| = 0 \) for \( p = 2 \).