$ x$ students took an exam with $ y$ problems. Each student solved a half of the problem and the number of person those who solved the problem is equal for each problem.For any two students exactly three problems could be solved by the two persons. Find all pairs of $ (x,\ y)$  satisfying this condition, then show an example to prove that it is possiible the case in terms of $ a,\ b$ as follows:
Note that every student solve $ (a)$ or don't solve $ (b)$.
1. **Identify the problem as a Balanced Incomplete Block Design (BIBD):**
   - We are given \( x \) students and \( y \) problems.
   - Each student solves half of the problems, so each student solves \( \frac{y}{2} \) problems.
   - Each problem is solved by the same number of students, denoted as \( k \).
   - Any two students solve exactly 3 problems in common.

2. **Define the parameters of the BIBD:**
   - \( v = x \) (number of students)
   - \( b = y \) (number of problems)
   - \( r = \frac{y}{2} \) (number of problems each student solves)
   - \( k \) (number of students solving each problem)
   - \( \lambda = 3 \) (number of problems solved in common by any two students)

3. **Use the BIBD relations:**
   - The first relation is \( vr = kb \):
     \[
     x \cdot \frac{y}{2} = k \cdot y \implies \frac{xy}{2} = ky \implies k = \frac{x}{2}
     \]
     This implies \( x \) must be even, so let \( x = 2x' \).

   - The second relation is \( (v-1)\lambda = r(k-1) \):
     \[
     (x-1) \cdot 3 = \frac{y}{2} \left(\frac{x}{2} - 1\right)
     \]
     Substituting \( x = 2x' \) and \( y = 2y' \):
     \[
     (2x' - 1) \cdot 3 = y' \left(x' - 1\right)
     \]
     Simplifying:
     \[
     3(2x' - 1) = y'(x' - 1)
     \]

4. **Solve the equation \( 3(2x' - 1) = y'(x' - 1) \):**
   - Since \( x' - 1 \) and \( 2x' - 1 \) are relatively prime, we consider two cases:
     - Case 1: \( x' = 2 \)
       \[
       3(2 \cdot 2 - 1) = y'(2 - 1) \implies 3 \cdot 3 = y' \cdot 1 \implies y' = 9
       \]
       Thus, \( (x, y) = (4, 18) \).

     - Case 2: \( x' = 4 \)
       \[
       3(2 \cdot 4 - 1) = y'(4 - 1) \implies 3 \cdot 7 = y' \cdot 3 \implies y' = 7
       \]
       Thus, \( (x, y) = (8, 14) \).

5. **Provide models for the solutions:**
   - For \( (x, y) = (4, 18) \):
     - Students: \( A, B, C, D \)
     - Problems: \( a, b, c, d, e, f \)
     - Problem sets: \( A \to abc, B \to ade, C \to bdf, D \to cef \)

   - For \( (x, y) = (8, 14) \):
     - Students: \( 1, 2, 3, 4, 5, 6, 7, 8 \)
     - Blocks (problems): 
       \[
       \{1, 3, 7, 8\}, \{1, 2, 4, 8\}, \{2, 3, 5, 8\}, \{3, 4, 6, 8\}, \{4, 5, 7, 8\}, \{1, 5, 6, 8\}, \{2, 6, 7, 8\}, \{1, 2, 3, 6\}, \{1, 2, 5, 7\}, \{1, 3, 4, 5\}, \{1, 4, 6, 7\}, \{2, 3, 4, 7\}, \{2, 4, 5, 6\}, \{3, 5, 6, 7\}
       \]

The final answer is \( \boxed{ (4, 18) } \) and \((8, 14)\).