Selection tests for the Balkan Mathematical Olympiad 2013

Let $d=\gcd (a,b)$ and $a=d{a}^{\prime }$, $b=d{b}^{\prime }$ with $a^{\prime },b^{\prime }$ two relatively prime positive integers. The equation becomes $$a^{\prime }{b}^{\prime }{d}^2+63=20a^{\prime }{b}^{\prime }d+12d$$ Therefore, $d$ divides $63$ and we have $$a^{\prime }{b}^{\prime }d+\frac{63}{d}=20a^{\prime }{b}^{\prime }+12$$ If $5<d<20$, then $a^{\prime }{b}^{\prime }d+\frac{63}{d}<20a^{\prime }{b}^{\prime }+12$, and this is impossible. Hence, $d=1,3,21$, or $63$.

1. If $d=1$, the equation is equivalent to $51=19a^{\prime }{b}^{\prime }$, which is impossible since $19$ does not divide $51$.

2. If $d=3$, the equation is equivalent to $9=17a^{\prime }{b}^{\prime }$, which is impossible since $17$ does not divide $9$.

3. If $d=21$, the equation is equivalent to $a^{\prime }{b}^{\prime }=9$. Because $a^{\prime },b^{\prime }$ are relatively prime positive integers, either $a^{\prime }=1,b^{\prime }=9$ or $a^{\prime }=9,b^{\prime }=1$. This leads to two solutions $(a,b)=(21,189)$ and $(a,b)=(189,21)$.

4. If $d=63$, the equation is equivalent to $43a^{\prime }{b}^{\prime }=11$, which is impossible since $43$ does not divide $11$.

Therefore, the equation has two solutions $(a,b)=(21,189)$ and $(a,b)=(189,21)$.