75th Romanian Mathematical Olympiad

We start noticing that $a\ge \frac{a}{(a,b)}>b$. Denote $d=(a,b)$. Then $a=dx$, $b=dy$, $[a,b]=dxy$ and $(x,y)=1$, $x>y$. This gives $x=dy+\frac{48}{xy}$ and $y=dx-\frac{312}{xy}$ (1). It follows that $xy\mid 48$ and $xy\mid 312$, hence $xy\mid (48,312)=24$. Relation (1) yields $xy(x-dy)=48$ and $xy(dx-y)=312$, whence $\frac{x-dy}{dx-y}=\frac{48}{312}=\frac{2}{13}$ and $x(13-2d)=y(13d-2)$, $13-2d>0$.

I. $d=1\Rightarrow 11x=11y\Rightarrow a=b$, false, since $a>b$. II. $d=2\Rightarrow 3x=8y$. Since $(x,y)=1$, we get $x=8$, $y=3$, hence $a=16$, $b=6$. III. $d=3\Rightarrow 7x=37y\Rightarrow x=37$, $y=7$ – false, since $xy\mid 24$. IV. $d=4\Rightarrow x=10y\Rightarrow x=10$, $y=1$ – false, since $xy\mid 24$. V. $d=5\Rightarrow x=21y\Rightarrow x=21$, $y=1$ – false, since $xy\mid 24$. VI. $d=6\Rightarrow x=76y\Rightarrow x=76$, $y=1$ – false, since $xy\mid 24$.