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Romanian Mathematical Olympiad - District Round

Algebra Difficulty 5.3 AIME, harder Prove it Romania

a) Solve the equation in the set of real numbers $[x]^2 - x = -0.99$ .

b) Show that, for every $a \le -1$ , the equation $[x]^2 - x = a$ has no real solutions.

Solution

a) The equation is written equivalently $[x]^2 - [x] = \{x\} - 0.99$ , thus $\{x\} - 0.99 \in \mathbb{Z}$ .
Since $0 \le \{x\} < 1$ , we deduce that $-0.99 \le \{x\} - 0.99 < 0.01$ therefore $\{x\} - 0.99 = 0$ so $\{x\} = 0.99$ . Also, from $[x]^2 - [x] = 0$ we deduce that $[x] = 0$ or $[x] = 1$ , so $x \in \{0.99; 1.99\}$ .

b) The equation is written equivalently $[x]^2 - [x] = \{x\} + a$ . We assume, by absurdity, that there is $a \le -1$ for which the equation has real solutions. Then, since $\{x\} < 1$ , $[x]^2 - [x] = \{x\} + a < 0$ .
Denoting $[x] = y \in \mathbb{Z}$ , we have $y(y - 1) < 0$ , so $y \in (0, 1)$ , absurd.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.