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I am have difficulty carrying out the following proof:

Let a,b and n be natural numbers. If a|n, b|n and g.c.d(a,b)=1 then prove that ab|n.

I understand that if a|n this implies n=ca for some natural number c and that b|n implies n=mb for some natural number m. I can also see that g.c.d(a,b)=1 implies that a and b share no common prime factors.

Any tips or hints on how to solve above would be greatly appreciated. Thank you.

Zach Jamison
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  • Thank you for letting us know. I will be sure to make a note of this. I was not aware that I was beholden to you and that you could order me around, but I guess I learn something new every day. Did you have a question, though? Because your post does not contain any. – Arturo Magidin Sep 25 '21 at 20:32
  • Apologies Anturo, I have edited the question and added more detail. – Zach Jamison Sep 25 '21 at 21:40
  • Hint: $a|n$, so $n=ak$ for some $k$. $b|n$, so $b|ak$. But $\gcd(a,b)=1$, so... – Arturo Magidin Sep 25 '21 at 21:59
  • Thank you very much for the hint. Since b|ak and gcd(a,b)=1 this implies that b|k, thus k=mb for some m. Substituting into n=ak gives n=abm, hence ab|n. I hope I am understanding this now! Thank you. – Zach Jamison Sep 25 '21 at 22:33

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