If $ a \mid bc $ then $\frac{a}{\gcd(a,b)} \mid c$?

Question

Prove or reject this statement:

If $ a \mid bc $ then $\displaystyle \frac{a}{\gcd(a,b)} \mid c$

Answer 1

Hint: If $a\mid bc$ and $(a,b)=1$ then $a\mid c$. Use this to show:

$$ \displaystyle a \mid bc \implies \frac{a}{\gcd(a,b)} \mid \frac{b}{\gcd(a,b)}\times c \implies \frac{a}{\gcd(a,b)}\mid c $$ Because $\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1$.

Answer 2

HINT:

Let the highest power of prime $p$ that divides $a,b,c$ be $A,B,C$ respectively

As $a|bc, A\le B+C\iff A-B\le C\ \ \ \ (1) $

Now, the highest power$(D)$ of prime $p$ that divides $\displaystyle \frac a{(a,b)}=A-$min$(A,B)$

If $A\ge B,D=A-B\le C$ by $(1)$

Else $D=A-A=0$ which is always $\le C$

Answer 3

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