proof - $\gcd(a, m) = \gcd(b, m) = 1 \implies \gcd(ab, m^2) =1$

Question

I have a proof and want to know if its correct.

Prove that $\gcd(a, m) = \gcd(b, m) = 1 \implies \gcd(ab, m^2) =1$

Proof:

$ax_0 + my_0 = 1$ and $bx_1 + my_1 = 1$
$ax_0 = 1 - my_0$ and $bx_1 = 1- my_1$ $$abx_0x_1 = 1 - my_0 - my_1 + my_0y_1$$ $$ab(x_0x_1) + m(y_0 + y_1) -m^2(y_0y_1) = 1$$ $$\gcd(ab, m, m^2) = 1 \implies \gcd(ab, m^2) = 1$$

Is my proof correct? Could you suggest a better way of proving the same? Any help would be appreciated.

Answer 1

Your proof is not correct: $\gcd(u,v,w) = 1$ does not necessarily imply $ \gcd(u,w) = 1$: take $u=2,v=3,w=4$.

You need to write $abx+m^2y=1$ but $ab(x_0x_1) + m(y_0 + y_1) -m^2(y_0y_1) = 1$ does not work for this.

Here is a proof along different lines:

Let $p$ be a prime divisor of both $ab$ and $m^2$.

Since $p$ divides $ab$, we must have that $p$ divides $a$ or $p$ divides $b$.

Since $p$ divides $m^2$, we must have that $p$ divides $m$.

Therefore, $p$ divides $a$ and $m$ or $b$ and $m$, which contradicts $\gcd(a, m) = \gcd(b, m) = 1 $.