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Problem 1541

National Olympiad, first round

Algebra Difficulty 6.0 Prove it European Mathematical Cup · North Macedonia

Which of the following claims are true, and which of them are false? If a fact is true you should prove it, if it isn't, find a counterexample.

a) Let $a, b, c$ be real numbers such that $a^{2013} + b^{2013} + c^{2013} = 0$ . Then $a^{2014} + b^{2014} + c^{2014} = 0$ .

b) Let $a, b, c$ be real numbers such that $a^{2014} + b^{2014} + c^{2014} = 0$ . Then $a^{2015} + b^{2015} + c^{2015} = 0$ .

c) Let $a, b, c$ be real numbers such that $a^{2013} + b^{2013} + c^{2013} = 0$ and $a^{2015} + b^{2015} + c^{2015} = 0$ . Then $a^{2014} + b^{2014} + c^{2014} = 0$ .

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.

Official solution

Firstly, we know that for every real number $x$ , $x^2 \ge 0$ holds. The key idea in this problem is that the expression $a^{2014} + b^{2014} + c^{2014}$ is a sum of squares (which are nonnegative numbers). Thus $a^{2014} + b^{2014} + c^{2014} = 0 \Leftrightarrow a = b = c = 0$ .

a) No: It is sufficient to find three real numbers whose sum equals $0$ , and then take their $2013$ th roots. For example $a = \sqrt[3]{1}$ , $b = \sqrt[3]{2}$ , $c = \sqrt[3]{-3}$ .

b) YES: From the key idea we conclude $a = b = c = 0$ and then we conclude $a^{2015} + b^{2015} + c^{2015} = 0 + 0 + 0 = 0$ .

c) NO: Again we have to find a counterexample, for instance $a = 1$ , $b = 0$ , $c = -1$ .

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