[b]Q12.[/b] Find all positive integers $P$ such that the sum and product of all its divisors are $2P$ and $P^2$, respectively.
1. Let \( P \) be a positive integer. We are given that the sum of all divisors of \( P \) is \( 2P \) and the product of all divisors of \( P \) is \( P^2 \).

2. Let \( \tau(P) \) denote the number of divisors of \( P \). The product of all divisors of \( P \) is known to be \( P^{\tau(P)/2} \). This is because each divisor \( d \) of \( P \) can be paired with \( P/d \), and the product of each pair is \( P \). Since there are \( \tau(P) \) divisors, there are \( \tau(P)/2 \) such pairs.

3. Given that the product of all divisors is \( P^2 \), we have:
   \[
   P^{\tau(P)/2} = P^2
   \]
   Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
   \[
   \tau(P)/2 = 2
   \]
   Therefore:
   \[
   \tau(P) = 4
   \]

4. The number of divisors \( \tau(P) = 4 \) implies that \( P \) can be either of the form \( q^3 \) (where \( q \) is a prime number) or \( P = qr \) (where \( q \) and \( r \) are distinct prime numbers).

5. We need to check both forms to see which one satisfies the condition that the sum of the divisors is \( 2P \).

6. **Case 1: \( P = q^3 \)**
   - The divisors of \( P \) are \( 1, q, q^2, q^3 \).
   - The sum of the divisors is:
     \[
     1 + q + q^2 + q^3
     \]
   - We need this sum to be \( 2P = 2q^3 \):
     \[
     1 + q + q^2 + q^3 = 2q^3
     \]
   - Rearranging, we get:
     \[
     1 + q + q^2 = q^3
     \]
   - This equation does not hold for any prime \( q \) because \( q^3 \) grows much faster than \( q^2 \).

7. **Case 2: \( P = qr \)**
   - The divisors of \( P \) are \( 1, q, r, qr \).
   - The sum of the divisors is:
     \[
     1 + q + r + qr
     \]
   - We need this sum to be \( 2P = 2qr \):
     \[
     1 + q + r + qr = 2qr
     \]
   - Rearranging, we get:
     \[
     1 + q + r = qr
     \]
   - Solving for \( q \) and \( r \), we can try small prime values:
     - Let \( q = 2 \) and \( r = 3 \):
       \[
       1 + 2 + 3 = 2 \cdot 3
       \]
       \[
       6 = 6
       \]
     - This holds true, so \( P = 2 \cdot 3 = 6 \).

8. Therefore, the only positive integer \( P \) that satisfies the given conditions is \( P = 6 \).

The final answer is \( \boxed{6} \)