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Problem 1445

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Algebra Difficulty 5.3 Prove it Ukrainian National Mathematical Olympiad - Third Round (Second Tour) · Ukraine

Numbers x1,x2,…,x2015x_1, x_2, \ldots, x_{2015} fulfill both equalities simultaneously:

x12014+x22014+…+x20152014=1x_1^{2014} + x_2^{2014} + \ldots + x_{2015}^{2014} = 1 and x12015+x22015+…+x20152015=−1x_1^{2015} + x_2^{2015} + \ldots + x_{2015}^{2015} = -1.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

From the first equation −1≤xi≤1-1 \le x_i \le 1, i=1,…,2015i = 1, \ldots, 2015, so for all ii: 0≤1+xi≤20 \le 1 + x_i \le 2. Add both equations and get
x12014(1+x1)+x22014(1+x2)+…+x20152014(1+x2015)=0. x_1^{2014}(1+x_1) + x_2^{2014}(1+x_2) + \ldots + x_{2015}^{2014}(1+x_{2015}) = 0.
Since every item xi2014(1+xi)≥0x_i^{2014}(1+x_i) \ge 0, then their sum equals zero if and only if every item equals zero. So all xi∈{−1,0}x_i \in \{-1, 0\}. From the first equation follows that exactly one variable is not null, from the second equation follows that this variable equals −1-1. Thus, the equation is satisfied only by such set of numbers: xi=−1x_i = -1, i=1,…,2015i = 1, \ldots, 2015, xj=0x_j = 0, j≠ij \ne i.

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