# 2003 Indonesia MO Problems/Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Find all real solutions of the equation $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.

[Note: For any real number $\alpha$, $\lfloor \alpha \rfloor$ is the largest integer less than or equal to $\alpha$, and $\lceil \alpha \rceil$ denote the smallest integer more than or equal to $\alpha$.]

## Solution

There are two cases to consider -- one where $x$ is positive and one where $x$ is negative.

For the positive case, if $x = \sqrt{a},$ then the equation results in $2a = 2003.$ Since the equation does not have an integral solution, $x \ne \sqrt{a}.$ If we let $\sqrt{a} < x < \sqrt{a+1}.$ That means $a + a + 1 = 2003,$ and solving the equation yields $a = 1001.$ For confirmation, $1001 \le \lfloor x^2 \rfloor < 1002$ and $1001 < \lceil x^2 \rceil \le 1002$, so $2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004$. Since $\lfloor x^2 \rfloor + \lceil x^2 \rceil$ can only have integral values, $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.

For the negative case, if $x = -\sqrt{a},$ then the equation results in $2a = 2003.$ This also does not have an integral solution, so $x \ne \sqrt{a}.$ If we let $-\sqrt{a+1} < x < -\sqrt{a}.$ That means $a+1 + a = 2003,$ and this equation also yields $a = 1001.$ For confirmation, $1001 < \lfloor x^2 \rfloor \le 1002$ and $1001 \le \lceil x^2 \rceil < 1002$, so $2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004$. Since $\lfloor x^2 \rfloor + \lceil x^2 \rceil$ can only have integral values, $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.

In interval notation, the solutions of the equation are $\boxed{(-\sqrt{1002}, -\sqrt{1001}) \cup (\sqrt{1001}, \sqrt{1002})}.$