Suppose gcd(a,b) = 1, prove gcd(na,nb) = n
Any help?
Hello qtpipi$\displaystyle n$ is a common divisor of $\displaystyle na$ and $\displaystyle nb$. Suppose now that $\displaystyle \gcd(na, nb) = m > n$. Then, since every common divisor of $\displaystyle na$ and $\displaystyle nb$ is a divisor of $\displaystyle \gcd(na, nb), n|m \Rightarrow np = m$, for some $\displaystyle p > 1$
$\displaystyle \Rightarrow np | na$
$\displaystyle \Rightarrow p|a$ and
$\displaystyle np|nb \Rightarrow p|b$.
So $\displaystyle p$ is a common divisor of $\displaystyle a$ and $\displaystyle b$. Contradiction, since $\displaystyle \gcd(a,b)=1$.
Therefore $\displaystyle gcd(na, nb) = n$.
Grandad