Simple greatest common divisor proof [closed]

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) $\gcd(b,c)=1$?

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

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• Well, what divisors do $a$ and $c$ have in common that $ab$ and $c^2$ don't? – fleablood Feb 14 '18 at 6:26
• Consider prime divisors of each argument. – Parcly Taxel Feb 14 '18 at 6:26
• The only divisor they have in common is 1? – Homaniac Feb 14 '18 at 6:27
• $\gcd(a,c) \mid \gcd(am,cn)\,$ for all integer $\,m,n\,$. Use that with $\,m=b, n=c,$. – dxiv Feb 14 '18 at 6:28
• 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 $\gcd(ab,c^2)=1$. – JMoravitz Feb 14 '18 at 6:32

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 $c^2$.

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

• 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. – tomasz Feb 14 '18 at 12:54

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

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