Showing $\gcd(a,b,c) = \gcd(a,\gcd(b,c))$ without using Bezout's lemma and only the definition of the gcd Let $a,b,c \in \mathbb{Z}$ where $(b,c)  \neq (0,0)$. I would like to show that
$$
\gcd(a,b,c) = \gcd(a,\gcd(b,c)).
$$
The definition of the greatest common divisor (gcd) we received in our class is
$$
\gcd(a_1,\dots,a_n) = \max\{ m \in \mathbb{N} \, | m \, | \, a_i \text{ for all }i=1,\dots,n \}.
$$
Now I was able to show that $\gcd(a,b,c) \geq \gcd(a,\gcd(b,c))$, so only the $\leq$-part is missing. I tried to solve it myself but could not do it, so I looked up for answers online (like this one). It seems that the crucial part is to show that
$$
g \, | \, b,c \quad \Longrightarrow \quad g \, | \, \gcd(b,c)
$$
However, all proofs I found for this use either Bezout's Identity or the least common multiple which we did not cover in our class yet. But now I still don't know how to solve my homework otherwise and find a short proof which only uses the definition of the gcd I wrote above.
Could you please help me here?
 A: Consider the prime factorizations of $a$ and $b$. If $a = p_1^{\alpha_1} \ldots p_k^{\alpha_k}$  and $b = p_1^{\beta_1} \ldots p_k^{\beta_k}$, where $\alpha_i,\beta_i$ are non-negative integers. Then we can show that $\gcd(a,b) = p^{\min\{\alpha_1,\beta_1\}} \ldots p^{\min\{\alpha_k,\beta_k\}}$.
Suppose $d = \gcd(a,b) =p_1^{r_1} \ldots p_k^{r_k}$. Since $d$ divides both $a$ and $b$,we have $r_i \le \alpha_i,\beta_i$. That is, $r_i \le \min\{\alpha_i,\beta_i \}$.
Let $l_i = \min\{\alpha_i,\beta_i\} -r_i $. Then $p_i^{l_i}d$ is also a common factor of $a$ and $b$ and $p_i^{l_i}d \ge d$. Since $d$ is the gcd, we must have $p_i^{l_i} =1$ , that is, $l_i=0$ and $r_i = \min\{\alpha_i,\beta_i\}$.
With that established, it follows that if $q$ is any common divisor of $a,b$, then $q$ must divide $d$. To see this, note that since $q$ divides $a$ and $b$, it's prime factors must be among the $p_i's$. So $q$ has the form $p_1^{s_1} \ldots p_k^{s_k}$ for none-negative integers $s_i$. Also we must have $s_i \le \alpha_i, \beta_i$, that is, $s_i \le \min\{\alpha_i, \beta_i \}=r_i$. It follows that $q$ divides $d$.

Now for your original problem.
Let $d = \gcd (a,b,c), e= \gcd (a, \gcd(b,c))$. If you can show that $d | e$ and $e|d$. Then you'd be done.
Since $d$ divides $b$ and $c$, $d$ must also divide $\gcd(b,c)$. $d$ divides $a$ too, so $d$ must divide e = $\gcd(a,\gcd(b,c))$
Now, $e$ divides $\gcd(b,c)$, so $e$ divides $b$ and $c$. $e$ also divides $a$, so $e$ must divide $d =\gcd(a,b,c)$.
