Balkan Mathematical Olympiad

Note that $2013$ is divisible by $11$, and that $x^5$ is congruent with $0$, $1$ or $-1$ modulo $11$. Since $4^y$ is congruent $4$, $5$, $9$, $3$ or $1$ modulo $11$, we get that $x^5=-1(\mathrm{mod}11)$, and $4^y=1(\mathrm{mod}11)$. Therefore $5|y$, and the given equation is of the form $a^5+b^5=2013^z=3^z\cdot 11^z\cdot 61^z$, where $a=x$ and $b=4^{\frac{y}{5}}$. Note that $2\nmid x$, and therefore $(a,b)=1$. Also $a^5+b^5=(a+b)(a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4)$, and by previous remark, we have $(a+b,a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4)=(a+b,5a^4)=(a+b,5)=1$.

Let us prove that $3\nmid a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4$. Suppose that this is not true, i.e. that $3|a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4$. If $3|ab$, then $3|a$ and $3|b$, a contradiction. Since $3^z|a^5+b^5$, this implies that $3^z|a+b$. Also, $$(a+b{)}^4>a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4>a+b,$$ And therefore $61^z|a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4$, and $11^z|a+b$. So $a+b\ge 33^z$, and hence $$(a+b{)}^2>61^z\ge a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4.$$ Let $a>b$ (we can similarly deal with the case $a<b$). Then $a^4-a^{3}b=a^3(a-b)\ge a^3$ and $a^2{b}^2-a{b}^3=(a-b)a{b}^2\ge a{b}^2$. So, $$a^4-a^{3}b+a^2{b}^2-a{b}^3+b^4\ge a^3+ab+b^4\ge 2a^2+ab+b^2\ge (a+b{)}^2,$$ a contradiction. Therefore, the given equation does not have any solution in positive integers.