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 Omielan Sep 27 at 0:03
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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 $c^2$.
If $r(ab)+s(c^2)=1$, then $(rb)a+(sc)c=1$ etc.
By contradiction suppose $\gcd(ab,c^2)=1$ and
$$\gcd(a,c)=d>1\implies d|a \quad d|c \implies d|\gcd(ab,c^2)$$