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Algebra Difficulty 7.5 National olympiad, round 2 Prove it Ukraine

a) Prove that the equality {x}+{2y}={y}+{2x}\{x\} + \{2y\} = \{y\} + \{2x\} for real numbers xx and yy implies {x}={y}\{x\} = \{y\}.

b) Do there exist integers n≥3n \ge 3 such that the equality {x}+{ny}={y}+{nx}\{x\} + \{ny\} = \{y\} + \{nx\} for real numbers xx and yy implies {x}={y}\{x\} = \{y\}?

(Here {a}=a−[a]\{a\} = a - [a], where [a][a] stands for the greatest integer that does not exceed the real number aa.)

Solution

a) Suppose that for some real numbers xx and yy the equality {x}+{2y}={y}+{2x}\{x\} + \{2y\} = \{y\} + \{2x\} holds. Let {x}=α\{x\} = \alpha, {y}=β\{y\} = \beta, α,β∈[0;1)\alpha, \beta \in [0;1). It suffices for α,β∈[0;1)\alpha, \beta \in [0;1) to consider the equality α+{2β}=β+{2α}\alpha + \{2\beta\} = \beta + \{2\alpha\}.

If 0≤γ<120 \le \gamma < \frac{1}{2}, then {2γ}=2γ\{2\gamma\} = 2\gamma, and for 12≤γ<1\frac{1}{2} \le \gamma < 1, {2γ}=2γ−1\{2\gamma\} = 2\gamma - 1. Note that for the cases 0≤α<12≤β<10 \le \alpha < \frac{1}{2} \le \beta < 1 and 0≤β<12≤α<10 \le \beta < \frac{1}{2} \le \alpha < 1, the equality α+{2β}=β+{2α}\alpha + \{2\beta\} = \beta + \{2\alpha\} cannot hold. If 0≤α,β<120 \le \alpha, \beta < \frac{1}{2} or 12≤α,β<1\frac{1}{2} \le \alpha, \beta < 1, then the equality α+{2β}=β+{2α}\alpha + \{2\beta\} = \beta + \{2\alpha\} takes the form α=β\alpha = \beta.

b) Answer: no, such integers do not exist. Take x=0x=0, y=1n−1y = \frac{1}{n-1}. Then {x}≠{y}\{x\} \neq \{y\}, but, as is easy to see, {x}+{ny}={y}+{nx}\{x\} + \{ny\} = \{y\} + \{nx\}.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.