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How can we prove for $a$, $b$ and $c$ positive integers that if

$$\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=d$$ then $$\gcd(a,b,c)=d$$

For $a$ and $b$ co-prime numbers, $\gcd(a,b)=1$, means that the pairs $(b,c)$ and $(a,c)$ are also co-prime numbers, then from the $\gcd$ commutativity and associativity:

$$\gcd(a,b,c)=\gcd(\gcd(a,b),c)=\gcd(\gcd(a,c),b)=\gcd(a,gcd(b,c))$$ $$\gcd(a,b,c)=\gcd(1,c)=\gcd(1,b)=\gcd(a,1)=1$$ This is correct because if $a$, $b$ and $c$ are mutually prime numbers their $\gcd$ is $1$, is this a special case only and we can found counterexamples for the above claim?

Daniel V
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  • More generally $,\color{#c00}{(a,b)\mid c}\Rightarrow (a,b,c) = ((a,b),c) = (a,b)(1,\color{#c00}{c/(a,b)}) = (a,b),$ by the linked gcd associative and distributive laws. – Bill Dubuque Mar 22 '22 at 18:20

3 Answers3

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Assume that $d = gcd(,)$ and $ \mid $. Let $e = gcd(a,b,c)$. Since $d$ divides $a$, $b$, and $c$, we must have $e \geq d$. But if we assume $e > d$ this contradicts our assumption that $d = gcd(a,b)$. Thus, $e = d$.

Marvin Adams
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Let us merely assume that $\gcd(a,b)=d$ and that $d$ divides $c.$ Clearly, $d$ is a common divisor of $a,b,$ and $c.$ Let $e$ be any other common divisor of $a,b,$ and $c.$ Since $e$ is a common divisor of $a$ and $b,$ it follows that $e$ divides $\gcd(a,b)=d.$ Hence $d$ is the greatest common divisor of $a,b,$ and $c.$

bof
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  • Thank you @bof, this means if $\gcd(a,b)=d$ and $d$ divides $c$ we can say that $\gcd(a,b,c)=d$? – Daniel V Nov 17 '17 at 08:12
  • I think so. I gave an argument. Is there something wrong with my argument? Do you have a counterexample? – bof Nov 17 '17 at 08:24
  • I agree with this $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=\gcd(d,c)=d$ since $d$ divides $c$ and obviously $d$ and that a common divisor cannot be greater than its inputs (i.e. greater than $d$) then $\gcd(c,d)=d$. – zwim Nov 17 '17 at 08:48
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If $gcd(a,b)=d\Rightarrow a=md,b=nd$ with $ m,n\in\mathbb{N}$ and so $gcd(m,n)=1$. If $gcd(m,n)=1$, then $gcd(n,m,k)=1$ for any $k\in\mathbb{N}$. Can you work out from here?

Erik T.
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QED
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  • Thx @Abishanka Saha, the case $\gcd(a,b,c)=1$ works for all pairs $(a,b)$, $(a,c)$ and $(b,c)$ being relatively prime numbers, But I'm trying to prove or find a counterexample for the case when $d>1$ – Daniel V Nov 17 '17 at 08:21