Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.5 National Olympiad Prove it Ukraine

Do there exist pairwise distinct positive integers $a$ , $b$ and $c$ , such that $\left\{ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right\} = 0$ ? The fractions are not necessarily irreducible.

Here $\{x\}$ denotes the difference between $x$ and the greatest integer that does not exceed $x$ , for example, $\left\{ \frac{7}{5} \right\} = \frac{2}{5}$ , $\left\{ \frac{2019}{3} \right\} = 0$ and $\left\{ \frac{2020}{3} \right\} = \frac{1}{3}$ .

Solution

A source for an example is the following equation: $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ . Clearly, irreducible fractions would not work, thus we will choose $\left\{ \frac{2}{12} \right\} = \left\{ \frac{1}{6} \right\} = \frac{1}{6}$ . Then $\left\{ \frac{12}{9} \right\} = \left\{ \frac{4}{3} \right\} = \frac{1}{3}$ and $\left\{ \frac{9}{2} \right\} = \frac{1}{2}$ . Thus, $\left\{ \frac{2}{12} \right\} + \left\{ \frac{12}{9} \right\} + \left\{ \frac{9}{2} \right\} = 1$ and $\{1\} = 0$ .

Another example is based on the equality $\frac{1}{2} + \frac{1}{2} + 0 = 1 \Rightarrow \left\{ \frac{1}{2} \right\} + \left\{ \frac{2}{4} \right\} + \left\{ \frac{4}{1} \right\} = 1$ .

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