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prove or disprove :

Suppose that $a$, $m$ and $n$ are positive integers such that $(m, n)=1$. If $m\mid a$ and $n\mid a$, then $mn\mid a$. I think it is true statement.

Let $m\mid a$ and $n\mid a$ then $mn\mid na$ and $nm\mid ma$ since $(m,n)=1$ then $mn\mid a$.

Ut is short answer and I am not sure if that is it complete answer?

Martin Sleziak
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dr.rise
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  • I wouldn't mark it correct unless you gave a more precise reason (than $(m,n)=1$) why $mn|a$ . – ancient mathematician Dec 10 '17 at 09:01
  • If $(m,n)=1$, then $(ma,na)=a$. if $mn\mid na$ and $mn\mid ma$, then $mn\mid a$. – 1ENİGMA1 Dec 10 '17 at 09:07
  • @ancientmathematician I see my mistake ,the correct that if m|na and since (m,n)=1 then m|a but which I did I think there is mistake any help with that please. – dr.rise Dec 10 '17 at 09:26
  • I think it is best to finish it off the way @José Carlos Santos does in his solution. – ancient mathematician Dec 10 '17 at 09:30
  • @ancientmathematician thank you so much – dr.rise Dec 10 '17 at 09:31
  • https://math.stackexchange.com/questions/173471/show-that-if-a-c-and-b-c-then-ab-c-when-a-is-coprime-to-b https://math.stackexchange.com/questions/194961/prove-if-am-and-bm-and-gcda-b-1-then-abm https://math.stackexchange.com/questions/407540/if-gcda-b-1-and-a-and-b-divide-c-then-so-does-ab – Martin Sleziak Dec 10 '17 at 10:26

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You prove correctly that $mn$ divides both $ma$ and $na$. Since $(m,n)=1$, there are $x,y\in\mathbb Z$ such that $xm+yn=1$ and therefore $xma+yna=a$. Since $mn$ divides both $ma$ and $na$, it follows from this that $mn\mid a$.

José Carlos Santos
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