# Given $\gcd(d,d')=1, d\mid n, d' \mid n$, show that $dd' \mid n$

Given $d,d'$ are in $\mathbb{Z} > 1$, and $\gcd(d,d')=1$, and $d \mid n$ and $d'\mid n$,

Show that $d\cdot d'\mid n$.

I pretty much have it but I think it could be made more clear. I have:

$d \mid n$ so $n=q*d$ and $d'\mid n$ so $n=q'*d'$

So $q*d*q'*d'=n^2$

So $(q*d*q'*d')/n=n$

So $(q*q'/n)*(d*d')=n$ ... so the only thing left to do is show that $q*q'/n$ must be an integer.

I think it has something to do with $\gcd(d,d')=1$ because $d$ and $d'$ cannot both divide an integer $n$ unless $n$ is a multiple of $d*d'$, but I'm not sure how to show this elegantly.

Thanks!

Hint: since $\gcd(d, d') = 1$, there exist $\alpha, \beta \in \mathbb{Z}$ such that $\alpha d + \beta d' =1$. Now multiply both sides by $n$, and use the fact that $d \mid n$ and $d' \mid n$. Feel free to comment if you need further clarification!