Simple greatest common divisor proof

Question

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

Answer 1

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

Answer 2

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

Answer 3

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 $c^2$.