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If $\gcd(a,b)=d_1$ and $\gcd(b,c)=d_2$. Here $d_1$ and $d_2$ are not $1$. In addition we have $\gcd(d_1,d_2)=1$. Can we say $\gcd(a,c)=1$ ? The question seems easy but I am not able to see whether it is true or not? If it is true please help me to prove this! Thank you!

Bill Dubuque
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    What if $d_1=d_2=1, a=c$? – Sergey Guminov Mar 28 '23 at 17:25
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    Let $a=6, b=10, c=15$. – Robert Shore Mar 28 '23 at 17:27
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    The hypotheses are equivalent to $(b,a)>1,\ (b,c)>1,\ (b,(a,c)) = 1,,$ i.e. $b$ is not coprime to both $a$ & $c$ but $b$ is coprime to their gcd. Now it is easy to see how to construct counterexamples, e.g. if $A,B,C$ are the sets of prime factors of $a,b,c$ then we need $B$ to meet $A$ & $C$ but not $A\cap C$, so now it's a simple set-theory problem (boolean algebra). – Bill Dubuque Mar 28 '23 at 17:46
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    Such set-theoretic reductions often greatly simplify number theory problems, e.g. see Stieltjes coprime form of Dirichlet's theorem on infinitely many primes in arithmetic progression – Bill Dubuque Mar 28 '23 at 17:49

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