Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024)

Without loss of generality, we can assume that $$P=\gcd (a,b)\cdot \text{lcm}(b,c)=\gcd (b,c)\cdot \text{lcm}(c,a)\cdot \gcd (c,a)\cdot \text{lcm}(a,b).$$ Using the well-known relation $\gcd (x,y)\cdot \text{lcm}(x,y)=xy$, we get: $$(abc{)}^2=(ab)\cdot (bc)\cdot (ca)=\gcd (a,b)\cdot \text{lcm}(a,b)\cdot \gcd (b,c)\cdot \text{lcm}(b,c)\cdot \gcd (c,a)\cdot \text{lcm}(c,a)=P^2,$$ which gives $abc=P$ after taking square roots. Now, we can use the same relation once more to get $$\gcd (a,b)\cdot \text{lcm}(b,c)=P=a\cdot (bc)=a\cdot \gcd (b,c)\cdot \text{lcm}(b,c),$$ which can be simplified to $\gcd (a,b)=a\cdot \gcd (b,c)$. Since greatest common divisor is a divisor, we see that $a\mid \gcd (a,b)\mid b$, hence $b$ is a multiple of $a$, as desired.