60th Belarusian Mathematical Olympiad

By condition it follows that one of the numbers is divisible by $5$. Moreover, exactly one of the numbers is divisible by $5$, otherwise $(a,b)$ and $90(a,b)$ are divisible by $25$, and so $160$ is divisible by $25$, a contradiction. Let without loss of generality $a\ne 5$, $b\ne 5$. Then $a=5c$, $(a,b)=(5c,b)=(c,b)$, and the equation can be rewritten as $bc=32+18(c,b)$. Since $bc$ and $18(c,b)$ are divisible by $(c,b)$, it follows that $(c,b)$ is a factor of $32$, i.e. $(c,b)=1,2,4,8,16,32$.

If $(c,b)\ge 4$, then $bc$ is divisible by $4^2=16$, so $18(c,b)=(bc-32)\ne 16$, and, therefore, $(c,b)\ne 8$. Thus $bc\ne 64$, i.e. $18(c,b)\ne 32$. Therefore, $(c,b)\ne 16$. If either $(c,b)=16$ or $(c,b)=32$, then $bc=16^2$, but $bc=32+18\cdot 16$ or $bc=32+18\cdot 64$, which are impossible. Hence $(c,b)\le 2$, i.e. is equal to either $1$ or $2$.

1) Let $(c,b)=1$. Then $bc=50$, and at least one of the numbers is odd. Since $b\ne 5$, we find that either $c=25$ and $b=2$ or $c=50$ and $b=1$. This gives the solutions of the initial equation: $(a,b)=(125,2)$ and $(a,b)=(250,1)$.

2) Now let $(c,b)=2$. Then $bc=32+36=68$. Set $b=2b_1$, $c=2c_1$, where $b_1$ and $c_1$ are coprime. Then $b_1{c}_1=17$. Hence either $c_1=1$ and $b_1=17$ or $c_1=17$ and $b_1=1$. This gives the solutions of the initial equation: $(a,b)=(10,34)$ and $(a,b)=(170,2)$.

Since the initial equation is symmetric with respect to $a$ and $b$, we have four more solutions: $(34;10)$, $(2;125)$, $(2;170)$, $(1;250)$.