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I'm having trouble with this simple proof. Any help would be appreciated. I don't really know where to start to try to conquer this problem.

Suppose $a|m$, $b|m$ and $\gcd(a,b) = 1$. Prove, without appealing to the fundamental theorem of arithmetic, that $ab|m$.

I know that $\gcd(a,b)=1$ means they are relatively prime. I also know that $a|m$ means $a=ms$ and that $b|m$ means $b=mt$ and lastly that $\gcd(a,b)=1$ means $1=as+bt$. I just have no idea what to do next. Any help would be appreciated. Thanks!

Mike Pierce
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WhatsAGuitar
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  • See this answer. – Bill Dubuque Aug 31 '15 at 22:03
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    You're making things difficult on yourself by being sloppy. e.g. $\gcd(a,b) = 1$ doesn't mean $1 = as+bt$. There is something similar you can say that's true, and those details are important. You speak as if you don't fully understand those details, and habitually ignoring them will make things hard on yourself! –  Sep 01 '15 at 03:13

2 Answers2

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Start from Bézout's identity: since $a,b$ are coprime, there exist integers $u,v\in \mathbf Z$ such that $ua+vb=1$, whence $$m=uam+vbm.$$ If $a\mid m$ and $b\mid m$, there exist integers $m_1,m_2$ such that $\;m=am_1,\enspace m=bm_2$. Combining with the above equation, we have: $$m=uabm_2+vbam_1=ab(um_2+vm_1).$$

Bernard
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The observation you make in your question isn't quite right: Given that $a\mid m$ and $b\mid m$, there exist integers $s$ and $t$ such that $$as=m\qquad\text{ and }\qquad bt=m.$$ Also, when you say that $\gcd(a,b)=1$ implies $1=as+bt$ you should be careful using the same letters $s$ and $t$ as before; they need not be the same numbers! In stead you can say there exist integers $x$ and $y$ such that $$ax+by=1.$$ Perhaps you can already take it from here? If not, here's a hint:

If $ax+by=1$ then $st=axst+byst=mxt+mys=m(xt+ys)$, so $m\mid st$.

  • Thanks for your help. I just realized the mistake I made in defining m for a and b. I just have one question: what made you go from $ax+by=1$ to $st=axst+byst$? Again, thank you for the help. – WhatsAGuitar Aug 31 '15 at 23:40
  • There are only so many ways to combine these three expressions. I thought whatever I would try should be symmetric in $a$ and $b$, and after some trial and error this turned out to work. –  Sep 01 '15 at 10:25