$\gcd(ab, c^2) = 1$, then $\gcd(a, c) = \gcd(b, c) = 1$

Question

Show that if $\gcd(ab, c^2)$ = 1, then $\gcd(a, c) = \gcd(b, c) = 1$

I already tried $$(a,c)(b,c) = ab,ac,cb,cc = ab, c(a,b,c)$$

then I'm stuck... no idea how to proceed on

Answer 1

Let $\gcd(a, c) = m$ and $\gcd(b, c) = n$.

We have $m|a$ and $m|c$, then $m|ab$ and $m|c^2$ $\therefore$ $m| \gcd(ab, c^2)$

The sames for $n$.

Since $m|1$ and $n|1$, $m=n=1$

$*$ Always suppose $\gcd > 0$

Answer 2

Since $$\gcd(ab,c^2)=1$$

$\exists x,y \in \mathbb{Z}$,

$$abx+c^2y=1$$

Hence we have $$a(bx)+c(cy)=1 \implies \gcd(a,c)=1$$

and $$b(ax)+c(cy)=1 \implies \gcd(b,c)=1$$

Answer 3

Put it in plain english.

$ab $ and $c^2$ have no factors in common. So none of the factors of $ab $ have factors in common with with the factors of $c^2$.

And that's it!

To put it in math talk: $\gcd (a,c)|a $ and $a|ab $ so $\gcd (a,c)|ab $. And $\gcd (a,c)|c $ and $c|c^2$ so $\gcd (a,c)|c^2$.

So $\gcd (a,c) $ is a common factor of $ab $ and $c^2$. But the greatest common divisor of $ab $ and $c^2$ is $1$. So $\gcd (a,b)\le \gcd(ab,c^2) =1$.

Same argument to show $\gcd (b,c)=1$.