Show that if g.c.d.(x,y) = g.c.d.(x,z) = 1 then g.c.d.(x,yz) = 1
Hello maths900$\displaystyle gcd(x, yz) > 1$ $\displaystyle \Rightarrow \exists$ a prime $\displaystyle p$ such that $\displaystyle p|x$ and $\displaystyle p |yz$
$\displaystyle \Rightarrow p |y$ or $\displaystyle p| z$
$\displaystyle \Rightarrow gcd(x, y) = p$ or $\displaystyle gcd(x,z) = p$
Contradiction.
$\displaystyle \Rightarrow gcd(x, yz) = 1$
Grandad
One important fact:
- If $\displaystyle (a,b) = k$ and $\displaystyle c \mid a $ and $\displaystyle c \mid b$, then $\displaystyle c \mid k \qquad {\color{red}\star}$
Basically, any common divisor of two numbers will divide the gcd of those two numbers.
___________
Let $\displaystyle d = (x,yz) \geq 1$.
$\displaystyle (x,y) = 1 \ \Leftrightarrow \ mx + ny = 1$
Multiply both sides by $\displaystyle z$: $\displaystyle mxz + nyz = z$
Since $\displaystyle d \mid x(mz)$ and $\displaystyle d \mid yz(n)$, then $\displaystyle d \mid z$. (Why?)
But this means $\displaystyle d \mid x$ and $\displaystyle d\mid z$ which implies $\displaystyle d \mid (x,z)$ because of $\displaystyle {\color{red}\star}$.
But $\displaystyle (x,z) = 1$. Can you conclude?