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I tried this: we have to prove $\gcd(b,c)=1$. $a$ is prime to $b$, then $\gcd(a,b)=1$. And at least one of $a$ or $b$ is prime. $c$ is a divisor of $a$ then $\gcd(c,a)=c$. And $a=kc$ where $k$ is an integer. It proves that $a$ is not prime. So $b$ is prime. Now since $b$ is prime, then $\gcd(b,c)=1$.

Is it correct?

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    $\begingroup$ 4 is prime to 9 but neither 4 nor 9 is a prime number. $\endgroup$ – Suzet Jun 11 '18 at 5:10
  • $\begingroup$ Then which method should I use? $\endgroup$ – user568963 Jun 11 '18 at 5:12
  • $\begingroup$ Basic definition of the gcd. Show that the only common divisors of b and c are 1 and -1 (I assume that you are working in $\mathbb{Z}$) $\endgroup$ – Suzet Jun 11 '18 at 5:15
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Let's $\gcd(c,b) = k$ then $k|b$ and $k|c$. Since $c|a$ we have $k|a$, so $k$ is a common divisor for $a$ and $b$. But since $a$ and $b$ are coprime, $k \le 1$ and so $k=1$.

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An alternative proof: You can take advantage of the Bezout property of the integers. That is gcd$(a,b)=1$ if and only if there exist integers $s,t$ with $as+bt=1$.

If gcd$(a,b)=1$, then there exists $s,t \in \mathbb{Z}$ with $as+bt=1$. If $c \mid a$, then $a=cd$. But then we have $(cd)s+bt=1$ which implies $c(ds)+bt=1$, so gcd$(c,b)=1$.

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