for the sake of all of us, let us assert and demonstrate:
the exists and its is unique
Consider the auxiliar set
as a piece of there exists .
But if then for some .
We are going to check that satisfy the for :
Applying the divisions’ algorithm to this pair we get
If then and .
This contradicts the choice for , then and the claim follows.
So is a common divisor of .
Now suppose that is a positive integer that , and for any other which divides implies .
Then, since divides both then .
Additionally implies and , hence
Then and . That is , but is the only possible.