Three real numbers satisfy the following statements: 
(1) the square of their sum equals to the sum their squares. 
(2) the product of the first two numbers is equal to the square of the third number. 

Find these numbers.
Let \( x, y, z \) be the three real numbers.

1. From the first condition, we have:
   \[
   (x + y + z)^2 = x^2 + y^2 + z^2
   \]
   Expanding the left-hand side, we get:
   \[
   x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = x^2 + y^2 + z^2
   \]
   Subtracting \( x^2 + y^2 + z^2 \) from both sides, we obtain:
   \[
   2xy + 2yz + 2zx = 0
   \]
   Dividing by 2, we get:
   \[
   xy + yz + zx = 0
   \]

2. From the second condition, we have:
   \[
   xy = z^2
   \]

3. Combining the two equations \( xy + yz + zx = 0 \) and \( xy = z^2 \), we substitute \( z^2 \) for \( xy \) in the first equation:
   \[
   z^2 + yz + zx = 0
   \]
   Factoring out \( z \), we get:
   \[
   z(x + y + z) = 0
   \]
   This implies that either \( z = 0 \) or \( x + y + z = 0 \).

4. Case 1: \( z = 0 \)
   - If \( z = 0 \), then from \( xy = z^2 \), we have:
     \[
     xy = 0
     \]
     This implies that either \( x = 0 \) or \( y = 0 \). Therefore, the possible solutions in this case are:
     \[
     (x, y, z) = (0, k, 0) \quad \text{or} \quad (x, y, z) = (l, 0, 0)
     \]
     where \( k \) and \( l \) are any real numbers.

5. Case 2: \( x + y + z = 0 \)
   - If \( x + y + z = 0 \), substituting \( z = -x - y \) into \( xy = z^2 \), we get:
     \[
     xy = (-x - y)^2
     \]
     Expanding the right-hand side, we have:
     \[
     xy = x^2 + 2xy + y^2
     \]
     Rearranging terms, we get:
     \[
     0 = x^2 + xy + y^2
     \]
     This is a quadratic equation in terms of \( x \) and \( y \). Solving for \( x \) in terms of \( y \), we use the quadratic formula:
     \[
     x = \frac{-y \pm \sqrt{y^2 - 4 \cdot 1 \cdot y^2}}{2 \cdot 1} = \frac{-y \pm \sqrt{-3y^2}}{2} = \frac{-y \pm y\sqrt{-3}}{2}
     \]
     Since \( \sqrt{-3} \) is not a real number, there are no real solutions in this case.

Therefore, the only solutions are from Case 1.

The final answer is \( \boxed{ (0, k, 0) } \) or \((l, 0, 0)\), where \( k \) and \( l \) are any real numbers.