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I must be missing something simple but how to show if $\gcd(ab,c^2)=1$ implies both i) $\gcd(a,c)=1$ and ii) $\gcd(b,c)=1$?

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closed as off-topic by user21820, postmortes, metamorphy, mick, John Omielan Sep 27 at 0:03

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    $\begingroup$ Well, what divisors do $a$ and $c$ have in common that $ab$ and $c^2$ don't? $\endgroup$ – fleablood Feb 14 '18 at 6:26
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    $\begingroup$ Consider prime divisors of each argument. $\endgroup$ – Parcly Taxel Feb 14 '18 at 6:26
  • $\begingroup$ The only divisor they have in common is 1? $\endgroup$ – Homaniac Feb 14 '18 at 6:27
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    $\begingroup$ $\gcd(a,c) \mid \gcd(am,cn)\,$ for all integer $\,m,n\,$. Use that with $\,m=b, n=c,$. $\endgroup$ – dxiv Feb 14 '18 at 6:28
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    $\begingroup$ Alternatively, suppose that $\gcd(a,c)=d>1$. Then $d$ is a divisor of $a$ and a divisor of $c$. It follows it is also a divisor of $ab$ and $c^2$, contradicting that $\gcd(ab,c^2)=1$. $\endgroup$ – JMoravitz Feb 14 '18 at 6:32
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Let $d = \gcd(a,c)$. Then $d|a$ and $d|c$ so there are integers $k,j$ such that $a=d*k$ and $c = d*j$.

The $ab = d*(bk)$ and $c^2= d*(jc)$. So $d$ divides both $ab$ and $c^2$. But $\gcd(ab, c^2) =1$ so the only (positive) common factor that $ab$ and $c^2$ have in common $1$. But they have $d$ in common.

So $d$ has to be $1$.

Same argument works to show that $\gcd(b,c)$ is a common divisor of $b$ and $c$ and thus a common divisor of $ab$ and $c^2$.

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If $r(ab)+s(c^2)=1$, then $(rb)a+(sc)c=1$ etc.

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    $\begingroup$ This is a correct argument, but it presupposes more knowledge than it should, judging by the level of the question. It is also less general (works in a smaller class of abstract rings) and less natural than the other one. $\endgroup$ – tomasz Feb 14 '18 at 12:54
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By contradiction suppose $\gcd(ab,c^2)=1$ and

$$\gcd(a,c)=d>1\implies d|a \quad d|c \implies d|\gcd(ab,c^2)$$

then $\gcd(a,c)=1$.

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    $\begingroup$ Final statement should read $\gcd(ab,c^2)\geq d$. It could still be greater. $\endgroup$ – JMoravitz Feb 14 '18 at 6:34
  • $\begingroup$ yes of course you are right I was just fixing it after posing, thanks $\endgroup$ – user Feb 14 '18 at 6:35
  • $\begingroup$ Then should I qualify d not equals 1 so the final part contradicts? $\endgroup$ – Homaniac Feb 14 '18 at 6:35
  • $\begingroup$ absolutely right @dxiv $\endgroup$ – user Feb 14 '18 at 6:38
  • $\begingroup$ @homaniac of course you assume $d>1$ and obtain a contradiction, since gcd always exists it must be 1 $\endgroup$ – user Feb 14 '18 at 6:47

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