Find all quadruples of real numbers $(a,b,c,d)$ such that the equalities 
\[X^2 + a X + b = (X-a)(X-c) \text{ and } X^2 + c X + d = (X-b)(X-d)\]
hold for all real numbers $X$.

[i]Morteza Saghafian, Iran[/i]
To find all quadruples of real numbers \((a, b, c, d)\) such that the equalities 
\[X^2 + aX + b = (X-a)(X-c) \quad \text{and} \quad X^2 + cX + d = (X-b)(X-d)\]
hold for all real numbers \(X\), we start by expanding and equating the coefficients of the polynomials.

1. **Expand and equate coefficients for the first equation:**
   \[X^2 + aX + b = (X-a)(X-c)\]
   Expanding the right-hand side:
   \[(X-a)(X-c) = X^2 - (a+c)X + ac\]
   Equating coefficients with the left-hand side:
   \[X^2 + aX + b = X^2 - (a+c)X + ac\]
   This gives us the system of equations:
   \[
   \begin{cases}
   a = -(a+c) \\
   b = ac
   \end{cases}
   \]
   Simplifying the first equation:
   \[a + a + c = 0 \implies 2a + c = 0 \implies c = -2a\]

2. **Expand and equate coefficients for the second equation:**
   \[X^2 + cX + d = (X-b)(X-d)\]
   Expanding the right-hand side:
   \[(X-b)(X-d) = X^2 - (b+d)X + bd\]
   Equating coefficients with the left-hand side:
   \[X^2 + cX + d = X^2 - (b+d)X + bd\]
   This gives us the system of equations:
   \[
   \begin{cases}
   c = -(b+d) \\
   d = bd
   \end{cases}
   \]
   Simplifying the second equation:
   \[d = bd \implies d(1 - b) = 0\]

3. **Analyze the cases for \(d\):**
   - **Case 1: \(d = 0\)**
     - From \(d = 0\), the second equation \(c = -(b+d)\) simplifies to \(c = -b\).
     - Substituting \(c = -2a\) from the first set of equations, we get:
       \[-2a = -b \implies b = 2a\]
     - Substituting \(b = 2a\) into \(b = ac\), we get:
       \[2a = a(-2a) \implies 2a = -2a^2 \implies a^2 = 1 \implies a = \pm 1\]
     - If \(a = 1\):
       \[
       \begin{cases}
       b = 2 \\
       c = -2 \\
       d = 0
       \end{cases}
       \]
       This gives the solution \((1, 2, -2, 0)\).
     - If \(a = -1\):
       \[
       \begin{cases}
       b = -2 \\
       c = 2 \\
       d = 0
       \end{cases}
       \]
       This gives the solution \((-1, -2, 2, 0)\).

   - **Case 2: \(b = 1\)**
     - From \(d(1 - b) = 0\), if \(b = 1\), then \(d\) can be any value.
     - Substituting \(b = 1\) into \(c = -(b+d)\), we get:
       \[c = -(1 + d)\]
     - Substituting \(b = 1\) into \(b = ac\), we get:
       \[1 = a(-1 - d) \implies a = -\frac{1}{1 + d}\]
     - Substituting \(c = -2a\) into \(c = -(1 + d)\), we get:
       \[-2a = -(1 + d) \implies 2a = 1 + d \implies a = \frac{1 + d}{2}\]
     - Equating the two expressions for \(a\):
       \[-\frac{1}{1 + d} = \frac{1 + d}{2} \implies -2 = (1 + d)^2 \implies (1 + d)^2 = -2\]
     - This is a contradiction since the square of a real number cannot be negative. Therefore, there are no solutions in this case.

4. **Combine the solutions:**
   The only valid solutions are from Case 1.

The final answer is \(\boxed{(-1, -2, 2, 0), (0, 0, 0, 0)}\)