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I found myself needing to prove this statement when attempting to prove that every common divisor divides the GCD. However, I just cannot think of a way to prove this! The logical framework I'm working in does not yet regard the Bezout Identity nor the Euclidean algorithm as true, it only assumes the following:

  1. $a$ divides $b$ if there exists an integer $n$ such that $b = an.$
  2. For every integer $m$ and $n \in N \backslash \{0\}$, there exist unique integers $q$ and $r$ with $m = qn + r$ and $0 ≤ r < n.$
  3. $\text{gcd}(a,b)$ is defined to be the largest common divisor of $a$ and $b$.

Any tips? I'm very stuck...

Baylee V
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  • Isn't $(2)$ the Euclidean algorithm? – qualcuno Mar 01 '21 at 11:59
  • If you can use prime factorization it is easy to prove. However I guess you cannot use that the ring which you are working in is a UFD? –  Mar 01 '21 at 12:08
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    @guidoar That's Euclid's division lemma (or theorem), but the OP is referring to Euclid's algorithm for finding the GCD (and thence the coefficients in Bézout's identity). – PM 2Ring Mar 01 '21 at 12:38
  • Well, you can show that every common divisor of $a$ & $b$ divides every expression of the form $ax+by$ for any $x$ & $y$. But then you're halfway to proving Bézout's identity. ;) – PM 2Ring Mar 01 '21 at 12:47

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