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GCD and LCM Inception

Source: Indian RMO 2013 Mumbai Region Problem 5

February 1, 2014
number theorygreatest common divisorleast common multiplenumber theory unsolved

Problem Statement

Let a1,b1,c1a_1,b_1,c_1 be natural numbers. We define a2=gcd⁡(b1,c1),        b2=gcd⁡(c1,a1),        c2=gcd⁡(a1,b1),a_2=\gcd(b_1,c_1),\,\,\,\,\,\,\,\,b_2=\gcd(c_1,a_1),\,\,\,\,\,\,\,\,c_2=\gcd(a_1,b_1), and a3=lcm⁡(b2,c2),        b3=lcm⁡(c2,a2),        c3=lcm⁡(a2,b2).a_3=\operatorname{lcm}(b_2,c_2),\,\,\,\,\,\,\,\,b_3=\operatorname{lcm}(c_2,a_2),\,\,\,\,\,\,\,\,c_3=\operatorname{lcm}(a_2,b_2). Show that gcd⁡(b3,c3)=a2\gcd(b_3,c_3)=a_2.
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