European Mathematical Cup

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 $a^{2014}+b^{2014}+c^{2014}=0\iff a=b=c=0$.

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 $a=1$, $b=0$, $c=-1$.