For all real numbers $x$, we denote by $\lfloor x \rfloor$ the largest integer that does not exceed $x$. Find all functions $f$ that are defined on the set of all real numbers, take real values, and satisfy the equality
\[f(x + y) = (-1)^{\lfloor y \rfloor} f(x) + (-1)^{\lfloor x \rfloor} f(y)\]
for all real numbers $x$ and $y$.

[i]Navneel Singhal, India[/i]
1. Let \( P(x, y) \) be the assertion of the given functional equation:
   \[
   P(x, y): f(x + y) = (-1)^{\lfloor y \rfloor} f(x) + (-1)^{\lfloor x \rfloor} f(y)
   \]

2. Consider \( P(x + y, z) \):
   \[
   f(x + y + z) = (-1)^{\lfloor z \rfloor} f(x + y) + (-1)^{\lfloor x + y \rfloor} f(z)
   \]

3. Substitute \( f(x + y) \) from \( P(x, y) \) into \( P(x + y, z) \):
   \[
   f(x + y + z) = (-1)^{\lfloor z \rfloor} \left[ (-1)^{\lfloor y \rfloor} f(x) + (-1)^{\lfloor x \rfloor} f(y) \right] + (-1)^{\lfloor x + y \rfloor} f(z)
   \]
   \[
   f(x + y + z) = (-1)^{\lfloor y \rfloor + \lfloor z \rfloor} f(x) + (-1)^{\lfloor x \rfloor + \lfloor z \rfloor} f(y) + (-1)^{\lfloor x + y \rfloor} f(z)
   \]

4. Now consider \( P(x, y + z) \):
   \[
   f(x + (y + z)) = (-1)^{\lfloor y + z \rfloor} f(x) + (-1)^{\lfloor x \rfloor} f(y + z)
   \]

5. Substitute \( f(y + z) \) from \( P(y, z) \) into \( P(x, y + z) \):
   \[
   f(x + y + z) = (-1)^{\lfloor y + z \rfloor} f(x) + (-1)^{\lfloor x \rfloor} \left[ (-1)^{\lfloor z \rfloor} f(y) + (-1)^{\lfloor y \rfloor} f(z) \right]
   \]
   \[
   f(x + y + z) = (-1)^{\lfloor y + z \rfloor} f(x) + (-1)^{\lfloor x \rfloor + \lfloor z \rfloor} f(y) + (-1)^{\lfloor x \rfloor + \lfloor y \rfloor} f(z)
   \]

6. Equate the two expressions for \( f(x + y + z) \):
   \[
   (-1)^{\lfloor y \rfloor + \lfloor z \rfloor} f(x) + (-1)^{\lfloor x \rfloor + \lfloor z \rfloor} f(y) + (-1)^{\lfloor x + y \rfloor} f(z) = (-1)^{\lfloor y + z \rfloor} f(x) + (-1)^{\lfloor x \rfloor + \lfloor z \rfloor} f(y) + (-1)^{\lfloor x \rfloor + \lfloor y \rfloor} f(z)
   \]

7. Simplify the equation:
   \[
   (-1)^{\lfloor y \rfloor + \lfloor z \rfloor} f(x) + (-1)^{\lfloor x + y \rfloor} f(z) = (-1)^{\lfloor y + z \rfloor} f(x) + (-1)^{\lfloor x \rfloor + \lfloor y \rfloor} f(z)
   \]

8. Rearrange terms:
   \[
   \left[ (-1)^{\lfloor y \rfloor + \lfloor z \rfloor} - (-1)^{\lfloor y + z \rfloor} \right] f(x) + \left[ (-1)^{\lfloor x + y \rfloor} - (-1)^{\lfloor x \rfloor + \lfloor y \rfloor} \right] f(z) = 0
   \]

9. To satisfy this equation for all \( x \) and \( z \), we must have:
   \[
   (-1)^{\lfloor y \rfloor + \lfloor z \rfloor} = (-1)^{\lfloor y + z \rfloor} \quad \text{and} \quad (-1)^{\lfloor x + y \rfloor} = (-1)^{\lfloor x \rfloor + \lfloor y \rfloor}
   \]

10. Consider \( y \) as a given real number and choose \( x \in (0, 1 - \{ y \}) \) and \( z = 1 - x \):
    \[
    \left[ (-1)^{\lfloor y \rfloor + \lfloor 1 - x \rfloor} - (-1)^{\lfloor y + 1 - x \rfloor} \right] f(x) + \left[ (-1)^{\lfloor x + y \rfloor} - (-1)^{\lfloor x \rfloor + \lfloor y \rfloor} \right] f(1 - x) = 0
    \]

11. Since \( x \in (0, 1 - \{ y \}) \), we have \( \lfloor x \rfloor = 0 \) and \( \lfloor 1 - x \rfloor = 0 \):
    \[
    \left[ (-1)^{\lfloor y \rfloor} - (-1)^{\lfloor y + 1 - x \rfloor} \right] f(x) + \left[ (-1)^{\lfloor x + y \rfloor} - (-1)^{\lfloor y \rfloor} \right] f(1 - x) = 0
    \]

12. This implies \( f(y) = 0 \) for all \( y \):
    \[
    f(y) = 0
    \]

13. Therefore, the only solution is:
    \[
    \boxed{f(x) = 0 \quad \forall x}
    \]