Find all quadruples of real numbers such that each of them is equal to the product of some two other numbers in the quadruple.
We need to find all quadruples of real numbers \((a, b, c, d)\) such that each number is equal to the product of some two other numbers in the quadruple. We will consider different cases based on the values of \(a, b, c,\) and \(d\).

1. **Case 1: \(a \ge b \ge c \ge d > 0\)**

    - Suppose \(a > 1\). Since \(a\) is the largest, \(a \neq bc\), \(a \neq bd\), and \(a \neq cd\). Therefore, \(a = ab\) or \(a = ac\). However, \(ab \le a\) and \(ac \le a\), which contradicts \(a > 1\). Thus, \(a \le 1\).
    
    - Suppose \(a = 1\). Then \(a\) must be the product of two numbers less than or equal to 1. The only way this is possible is if \(a = b = c = d = 1\). This gives the solution \((1, 1, 1, 1)\).
    
    - Suppose \(a < 1\). Then \(a\) cannot be the product of any two numbers in the quadruple because \(a\) is the largest and all products would be less than \(a\). This is a contradiction.

2. **Case 2: \(a \ge b \ge c > 0 > d\)**

    - Here, \(d\) must be the product of two positive numbers, which is impossible since \(d\) is negative. This case leads to a contradiction.

3. **Case 3: \(a \ge b > 0 \ge c \ge d\)**

    - Since \(a\) is positive and the largest, \(a\) must be the product of two negative numbers. The only way this is possible is if \(a = cd\). Similarly, \(b = cd\). Thus, \(a = b = cd\).
    
    - Symmetrically, \(c = d = ab\). Therefore, \(\sqrt{cd} = \sqrt{c^2} = c = ab = a^2 = (cd)^2\). This implies \((cd)^3 = 1\), so \(cd = \pm 1\). Since \(cd > 0\), \(cd = 1\), meaning \(c^2 = d^2 = 1\). Thus, \(c = d = -1\) and \(a = b = 1\). This gives the solution \((1, 1, -1, -1)\).

4. **Case 4: \(a > 0 > b \ge c \ge d\)**

    - Suppose \(a > 1\). Then \(|ad| \ge |ac| > |c| \ge |b|\), so \(b \neq ac\) and \(b \neq ad\). Thus, \(b = cd > 0 > b\), which is a contradiction.
    
    - Suppose \(a < 1\). Then \(|ab| \le |ac| < |c| \le |d|\), so \(d \neq ab\) and \(d \neq ac\). Thus, \(d = bc > 0 > d\), which is a contradiction.
    
    - Suppose \(a = 1\). Then \(a\) must be the product of two numbers, and since \(a = 1\), we must have \(b = c = d = -1\). This gives the solution \((1, -1, -1, -1)\).

5. **Case 5: \(0 > a \ge b \ge c \ge d\)**

    - This case is impossible because \(a, b, c, d\) are all negative, contradicting any equality between one of them (negative) and the product of any two (positive).

Overall, the solutions are \((0, 0, 0, 0)\), \((1, 1, 1, 1)\), \((1, 1, -1, -1)\), and \((1, -1, -1, -1)\) along with their permutations.

The final answer is \( \boxed{ (0, 0, 0, 0) } \), \((1, 1, 1, 1)\), \((1, 1, -1, -1)\), and \((1, -1, -1, -1)\) along with their permutations.