Question

The only common divisors of two compliment

Answer 1

Answer:

The following is more general than for the integers, and therefore simpler (but longer than a proof using unique factorisation without proving it; here we start from scrap).

Let R be an integral domain, where d=gcd(a,b) is defined to mean that d∣a,b and d′∣a,b⟹d′∣d for all d′∈R, while m=lcm(a,b) is defined to mean that a,b∣m and a,b∣m′⟹m∣m′ for all m′∈R (in both cases it is not implied that gcd(a,b) or lcm(a,b) always exist, and if they do they are only unique up to multiplication by invertible elements; as a consequence in this setting the equality gcd(a,b)×lcm(a,b)=ab can only be asserted up to such multiplication, or for properly chosen values on the left hand side).