Romanian Mathematical Olympiad

a) $(\lfloor x \rfloor + \{x\})^2 + 4\lfloor x \rfloor = \{x\}^2$ yields $\lfloor x \rfloor^2 + 2\lfloor x \rfloor\{x\} + 4\lfloor x \rfloor = 0$, (*).
If $x \ge 0$, then $\lfloor x \rfloor = 0$, thus any $x \in [0, 1)$ is a solution.
If $x < 0$, equality (*) leads to $\{x\} = -\frac{\lfloor x \rfloor^2 + 4\lfloor x \rfloor}{2\lfloor x \rfloor} = -\frac{\lfloor x \rfloor + 4}{2}$. Since $\{x\} \in [0, 1)$, we conclude that $\lfloor x \rfloor \in \{-5, -4\}$, thus obtaining the solutions $x_1 = -\frac{9}{2}$, $x_2 = -4$, which verify the equality.
Therefore, $x \in \{-\frac{9}{2}, -4\} \cup [0, 1)$.

b) For $y \in (0, 1)$ we have $A(y) = y^2 \in (0, 1)$, $A(1) = 5$, and for $y \in [1, 2)$ we have $A(y) = y^2 + 4 \in (5, 8)$, thus $A(y)$ is not a perfect square for $y \in (0, 2)$.
For $y \ge 2$, the condition from the statement leads to $y^2 \in \mathbb{N}$, therefore $y = \sqrt{m}$, with $m \in \mathbb{N}$, $m \ge 4$, leading to $m + 4\lfloor\sqrt{m}\rfloor = p^2$, with $p \in \mathbb{N}$.
If $k = \lfloor\sqrt{m}\rfloor$, then $k \ge 2$ and $k^2 \le m < (k+1)^2$, hence $m + 4k = p^2$.
Using the fact that $k^2 + 4k \le p^2 < (k+1)^2 + 4k = k^2 + 6k + 1 < (k+3)^2$, it follows that (1) $p^2 = (k+1)^2 = 1$, for $k=0$ or (2) $p^2 = 9$ for $k=1$, values that do not verify the equality in the statement, or (3) $p^2 = (k+2)^2$, for $k \ge 2$.

Therefore, $p = k + 2$ and $m = (k + 2)^2 - 4k = k^2 + 4$. Thus, the numbers that verify the equality in the statement are $y = \sqrt{k^2 + 4}$, $k \in \mathbb{N}$, $k \ge 2$.