Bezout identity corollary generalization

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SUMMARY

The discussion centers on proving a corollary of Bezout's Identity, specifically that if A1, A2, ..., Ar are factors of m and (Ai, Aj) = 1 for all i ≠ j, then the product A1A2...Ar is also a factor of m. The approach suggested involves using mathematical induction (MI) on the number of pairs and leveraging associativity to express the product of several integers as a product of two integers. The conversation highlights the importance of starting with a single factor and building upon it to demonstrate the validity of the corollary.

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  • Understanding of Bezout's Identity
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Mathematicians, students studying number theory, and anyone interested in proofs related to Bezout's Identity and factorization concepts.

davon806
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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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