0
$\begingroup$

everyone I'm stuck on this proof. I know that $\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)$ but I don't know how I use $\gcd(b,c)=1$ to get that $\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)$

$\endgroup$

marked as duplicate by Juniven, Benjamin Dickman, Leucippus, Claude Leibovici, user91500 Apr 2 '17 at 9:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

We can write $a = p_1^{\alpha_1} \cdots p_n^{\alpha_n}$, $b = p_1^{\beta_1} \cdots p_n^{\beta_n}$ and $c = p_1^{\gamma_1}\cdots p_n^{\gamma_n}$, where $p_1, \dots, p_n$ are distinct primes and $\alpha_i, \beta_i, \gamma_i \ge 0$ for each $i$. Since $\gcd(b, c)$ = 1, what do we know about the value of $\min\{\beta_i, \gamma_i\}$ for each $i$? Use the prime factorizations to work out $\gcd(a, bc)$, $\gcd(a, b)$, and $\gcd(a, c)$ in terms of $p_1, \dots, p_n$. From there, use the value of $\min\{\beta_i, \gamma_i\}$ to argue that $\gcd(a, bc) = \gcd(a, b)\cdot \gcd(a, c)$.

$\endgroup$
  • $\begingroup$ It's not necessary to use FTA. $\endgroup$ – Xam Apr 2 '17 at 21:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.