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Question

Let a1,b1,c1 be natural numbers. We define
a2=gcd(b1,c1),b2=gcd(c1,a1),c2=gcd(a1,b1),
and
a3=lcm(b2,c2),b3=lcm(c2,a2),c3=lcm(a2,b2).
Show that gcd(b3,c3)=a2.

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Solution

For a prime p and a natural number n we shall denote by vp(n) the power of p dividing n.
Then it is enough to show that vp(a2)=vp(gcd(b3,c3)) for all primes p.
Let p be a prime and let α=vp(a1),β=vp(b1) and γ=vp(c1).
Bysymmetry, we assume that αβγ.
vp(a2)=min{β,γ} = β and similarly vp(b2)=vp(c2)=α.
vp(b3)=max{α,β} = β and similarly vp(c3)=max{α,β} = β.
vp(gcd(b3,c3))=vp(a2)=β

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