Product of Coprime Factors

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Theorem

Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime.

Let both $a$ and $b$ be divisors of $c$.


Then $ab$ is also a divisor of $c$.

That is:

$a \perp b \land a \backslash c \land b \backslash c \implies a b \backslash c$


Proof

We have:

  • $a \backslash c \implies \exists r \in \Z: c = a r$
  • $b \backslash c \implies \exists s \in \Z: c = b s$


So:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle a\) \(\perp\) \(\displaystyle b\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \exists m, n \in \Z: m a + n b\) \(=\) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Integer Combination of Coprime Integers          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle c m a + c n b\) \(=\) \(\displaystyle c\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle b s m a + a r n b\) \(=\) \(\displaystyle c\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle a b \left({s m + r n}\right)\) \(=\) \(\displaystyle c\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle a b\) \(\backslash\) \(\displaystyle c\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


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