60th Belarusian Mathematical Olympiad

Let $[a,b]=x$, and $(a,b)=y$. It is known (and easy to prove) that $ab=[a,b]\cdot (a,b)$, so $ab=xy$. Then the initial equation can be rewritten in the form $$xy=300+7x+5y\iff xy-7x-5y+35=335\iff x(y-7)-5(y-7)=335\iff (x-5)(y-7)=5\cdot 67.$$ Factors on the left-hand side are nonnegative integers, and since $x=[a,b]\ge (a,b)=y$, we have $(x-5)>(y-7)$. So we have two possibilities.

1) $x-5=67$ and $y-7=5$. Then $x=72$ and $y=12$. It means that $a=12n$, $b=12m$ for some relatively prime $n,m$, and $ab=xy=72\cdot 12$, i.e. $12n\cdot 12m=72\cdot 12$, thus $nm=6$. Taking into account that $a\le b$ and so $n\le m$, we obtain $n=1,m=6$. Therefore $a=12,b=12\cdot 6=72$, or $n=2,m=3$, so $a=12\cdot 2=24$, $b=12\cdot 3=36$.

2) $x-5=335$ and $y-7=1$. Then $x=340$ and $y=8$. But it is impossible since $[a,b]=x$ must be divisible by $y=(a,b)$, but $340\nmid 8$.

Therefore, we have two required pairs $(12,72)$, $(24,36)$.