gcd(a,b) x lcm(b,c), gcd(b,c)x lcm(c,a), ,gcd(c,a) x lcm(a,b)

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gcd(a,b) x lcm(b,c), gcd(b,c)x lcm(c,a), ,gcd(c,a) x lcm(a,b)

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Source: 2024 Czech and Slovak Olympiad III A p1

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#1 h May 18, 2024, 7:13 AM

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Let $a, b, c$ be positive integers such that one of the values $gcd(a,b) \cdot lcm(b,c), \,\,\,\, gcd(b,c)\cdot lcm(c,a), \,\,\,\, gcd(c,a)-\cdot lcm(a,b)$ is equal to the product of the remaining two. Prove that one of the numbers $a, b, c$ is a multiple of another of them.

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#2 h May 18, 2024, 12:32 PM

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since the equations are symmetric

WLOG (1) eqn = product of (2)*3
we have also
$a*b = \gcd(a,b)*lcm(a,b) \forall a,b$
so rewrite as
$\gcd(a,b)*\frac{b*c}{\gcd(b,c)} =c*a\gcd(b,c)*\frac{a*b}{\gcd(a,b)}$
this implies
$\gcd(a,b) = a*\gcd(b,c)$
$\gcd(a,b) =a*\gcd(b,c)\leq a$
$\gcd(b,c) = 1$ or we could said relative prime
$\gcd(a,b)$ indeed also exactly $a$ implies that b exactly multiples of a

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#3 h May 18, 2024, 1:07 PM

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Another way to solve
Let $\gcd(a,b) \cdot lcm(b,c) = \gcd(b,c)\cdot lcm(c,a) \gcd(c,a)\cdot lcm(a,b)$ or
$(\gcd(a,b) \cdot lcm(b,c))^2 = \gcd(a,b) \cdot lcm(b,c) \gcd(b,c)\cdot lcm(c,a) \gcd(c,a)\cdot lcm(a,b)=a^2b^2c^2 \to \gcd(a,b) \cdot lcm(b,c)=abc\geq a \cdot lcm(b,c) \to \gcd(a,b) \geq a \to a|b$

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#4 h Mar 30, 2025, 11:22 AM

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WLOG assume that first term is product of last two. If we use $gcd(m,n)*lcm(m,n)=mn$ our equation will be simplified as: $gcd(a,b)^2=ac*gcd(a,c)^2$ .
From here we can conclude that $gcd(a,c)|gcd(a,b)$ . Let $x=gcd(b,c)$ and $b=Bx$ , $c=Cx$ and $a=Ax$ . Then we have that: $gcd(A,B)^2=x^2AC$ .
It also implies that $x|gcd(A,B)$ . Let $gcd(A,B)=xd$ , and $A=dxn, B=dxm$ . Then our equation simply implies: $d=Cxm$ , which implies that $Cx|dx^2m$ , thus $c|a$ , and we are done.

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