floor

$\begin{aligned} \left \lfloor x^2 \right \rfloor - \left \lfloor x \right \rfloor ^2 & = 1999 \\ \left \lfloor (\lfloor x \rfloor + \{x\})^2 \right \rfloor - \left \lfloor x \right \rfloor ^2 & = 1999 \\ \left \lfloor \lfloor x \rfloor^2 + 2\lfloor x \rfloor \{x\} + \{x\}^2 \right \rfloor - \left \lfloor x \right \rfloor ^2 & = 1999 \\ \left \lfloor 2\lfloor x \rfloor \{x\} + \{x\}^2 \right \rfloor & = 1999 \\ \implies 2\lfloor x \rfloor \{x\} + \{x\}^2 & \ge 1999 \end{aligned}$

This implies that the smallest $\lfloor x \rfloor = \left \lceil \dfrac {1999}2 \right \rceil = 1000$ . Then, $\{x\}$ is given by:

$\begin{aligned} \{x\}^2 + 2000 \{x\} - 1999 = 0 \\ \implies \{x\} & = \frac {-2000 + \sqrt{2000^2+4\cdot 1999}}2 & \small \color{#3D99F6} \text{Note that }0 \le \{x\} < 1 \\ & = - 1000 + \sqrt{1001999} \\ \implies x & = \lfloor x \rfloor + \{x\} \\ & = 1000 - 1000 + \sqrt{1001999} \\ & = \sqrt{1001999} \end{aligned}$

$\implies y = \boxed{1001999}$

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Notations:

• $\{\cdot\}$ denotes the fractional part function .
• $\lceil \cdot \rceil$ denotes the ceiling function .