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max a+b+c given gcd(a, b, c)=1, gcd(a, b+c)>1, gcd(b, c+a)>1, gcd(c, a+b)>1

Source: 2015 Grand Duchy of Lithuania, Mathematical Contest p4 (Baltic Way TST)

October 3, 2020
number theorygreatest common divisormax

Problem Statement

We denote by gcd (...) the greatest common divisor of the numbers in (...). (For example, gcd(4,6,8)=2(4, 6, 8)=2 and gcd (12,15)=3(12, 15)=3.) Suppose that positive integers a,b,ca, b, c satisfy the following four conditions: ∙\bullet gcd (a,b,c)=1(a, b, c)=1, ∙\bullet gcd (a,b+c)>1(a, b + c)>1, ∙\bullet gcd (b,c+a)>1(b, c + a)>1, ∙\bullet gcd (c,a+b)>1(c, a + b)>1. a) Is it possible that a+b+c=2015a + b + c = 2015? b) Determine the minimum possible value that the sum a+b+ca+ b+ c can take.
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