Bezout identity corollary generalization

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The discussion revolves around proving a corollary of Bezout's Identity, specifically that if A1,...,Ar are factors of m and each pair (Ai,Aj) is coprime, then the product A1A2...Ar is also a factor of m. The original poster is seeking assistance with this proof, suggesting the use of mathematical induction (MI) and associativity in their approach. A participant proposes starting with one factor A1 and exploring the implications of finding another factor A2 that is coprime to A1. The conversation emphasizes building upon these relationships to demonstrate the overall factorization. The thread highlights the challenge of the proof and invites alternative methods for resolution.
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OP warned about not including an attempt at a solution

Homework Statement


Hi,
I have been trying to prove one of the corollaries of the Bezout's Identity in the general form.Unfortunately,I can't figure it out by myself.I hope someone could solve the problem.

If A1,...,Ar are all factors of m and (Ai,Aj) = 1 for all i =/= j,then A1A2...Ar is a factor of m

The writer is asking for a proof using MI on the number of pairs (and associativity to write a product of several intergers as a product of 2 integers, e.g. abcd = (abc)d.However,I am welcome to any different ways of approach.

Homework Equations



Bezout's identity

ax + by = 1

The Attempt at a Solution


I really have no idea on this,and my work was a mess,so I am not going to post it.
 
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What have you already tried and where did it break down?

I would start by assuming you only have A1.
m/A1 = X1.
If there is a factor of X1, call it A2, that has the property (A1, A2) = 1, then x*A2=X1, so A1*A2*x = A1*X1 = m.
Build from there.

If at any point your X becomes 1, you have a full set of factors.
 

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