Junior Macedonian Mathematical Olympiad

For $x=2k$, $x^4=16k^4\equiv 0(\mathrm{mod}16)$. For $x=2k+1$, $x^4-1=8k(k+1)(2k^2+2k+1)\equiv 0(\mathrm{mod}16)$, i.e. $x^4\equiv 1(\mathrm{mod}16)$. Since $2016^3-1\equiv 15(\mathrm{mod}16)$, and the sum of the numbers on the left-hand side never gives a remainder $15$ when divided by $16$, it follows that the given equation has no solution in the integers.