Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Croatia

Find all real numbers $x$ such that
$\left\lfloor \frac{x^2 + 1}{x + 2} \right\rfloor + \left\lfloor \frac{x - 1}{2} \right\rfloor = \frac{x(3x + 1)}{2(x + 2)}$
For a real number $t$ , $\lfloor t \rfloor$ denotes the largest integer not greater than $t$ . For example, if $t = 3.14$ , then $\lfloor t \rfloor = 3$ .

Solution

Note that
$\frac{x^2 + 1}{x + 2} + \frac{x - 1}{2} = \frac{x(3x + 1)}{2(x + 2)}.$
Since $\lfloor t \rfloor \le t$ holds for all real numbers $t$ , and the equality is attained if and only if $t$ is an integer, it follows that both
$\frac{x^2 + 1}{x + 2} \quad \text{and} \quad \frac{x - 1}{2}$
must be integers.
The latter is an integer if and only if $x$ is an odd integer, hence considering
$\frac{x^2 + 1}{x + 2} = x - 2 + \frac{5}{x + 2}$
we get $x + 2 \in \{1, -1, 5, -5\}$ , i.e. $x \in \{-1, -3, 3, -7\}$ .

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