Mongolian Mathematical Olympiad

$\frac{3}{2014}=\frac{1}{p}+\frac{1}{q}\Rightarrow 3pq=2014(p+q)=2\cdot 19\cdot 53(p+q).$

i) Consider the case $2\cdot 19\cdot 53\mid p$. Setting $p=2\cdot 19\cdot 53r$ we get $3rq=2\cdot 19\cdot 53r+q\Rightarrow q=\frac{2\cdot 19\cdot 53r}{3r-1}$. Since $q\in \mathbb{N}$, $3r-1\mid 2\cdot 19\cdot 53r$, $(r,3r-1)=1\Rightarrow 3r-1$ may take values 2, 38, 53, 1007. Therefore $$\begin{array}{rlrl}3r-1& =2\Rightarrow & r& =1,\\ 3r-1& =38\Rightarrow & r& =13,\\ 3r-1& =53\Rightarrow & r& =18,\\ 3r-1& =1007\Rightarrow & r& =336\end{array}$$ and corresponding 4 pairs $(p,q)$ are $(2014,1007)$, $(26182,689)$, $(36252,684)$, $(676704,672)$.

ii) Consider the case $2\cdot 19\nmid p$ and $53\nmid p$. Setting $p=2\cdot 19\cdot r$ we get $3\cdot 2\cdot 19\cdot rq=2\cdot 19\cdot 53(2\cdot 19r+q)$. Let $q=53s$. Consequently, $s=\frac{2\cdot 19r}{3r-53}$, $(r,3r-53)=1\Rightarrow 3r-53\mid 2\cdot 19r$. Therefore $$\begin{array}{rlrl}3r-53& =1\Rightarrow & r& =18,\\ 3r-53& =19\Rightarrow & r& =24\end{array}$$

iii) Consider the case $2\cdot 53\nmid p$ and $19\nmid p$. From here follows $\Rightarrow p=2\cdot 53r$ and setting $q=19s$ we get $s=\frac{2\cdot 53r}{3r-19}$. It implies $3r-19\mid 2\cdot 53r$. $3r-19=2\Rightarrow r=7$ $3r-19=53\Rightarrow r=24.$ If $r=7$ then $p=2\cdot 7\cdot 53$, $q=7\cdot 19\cdot 53\Rightarrow (p,q)=(742,7049)$ If $r=24$ then $p=2\cdot 24\cdot 53$, $q=19\cdot 48\Rightarrow (p,q)=(2544,912)$

iv) Consider the case $19\cdot 53\nmid p$ and $2\nmid p$. Setting $p=19\cdot 53r$, $q=2s$ we get $s=\frac{19\cdot 53r}{3r-2}$. Therefore $3r-2\mid 19\cdot 53r$. $3r-2=1\Rightarrow r=1;$ $3r-2=19\Rightarrow r=7$. Corresponding pairs are $(p,q)=(1007,2014);(7049,742)$.

Thus there are 6 possibilities: $\frac{3}{2014}=\frac{1}{2014}+\frac{1}{1007}=\frac{1}{26182}+\frac{1}{689}=$ $$\frac{1}{676704}+\frac{1}{672}=\frac{1}{36252}+\frac{1}{684}=\frac{1}{912}+\frac{1}{2544}=\frac{1}{7049}+\frac{1}{742}.$$