Prove that if $n_1$ and $n_2$ divide $n$ and $\gcd(n_1,n_2) = 1$, then $n_1 n_2$ divides $n$.
I start by letting $n = n_1 k$ and $n = n_2 h$, and I know I'm trying to show that $n = q n_1 n_2$ but I can't figure a way to show this.
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Can you take prime decompositions? – Luiz Cordeiro Aug 08 '16 at 19:09
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1Hint try and use the fact that since $gcd(n_1,n_2) =1$ there exist $a,b \in \mathbb{Z}$ such that $1= an_1 + bn_2$ – Nex Aug 08 '16 at 19:12
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@Nex : Note the use of \gcd in MathJax, as in my edit to the question. $\qquad$ – Michael Hardy Aug 08 '16 at 19:15
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@MichaelHardy Thanks. – Nex Aug 08 '16 at 19:17
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Let $x$ and $y$ be such that $1=xn_1+yn_2$. Multiplying both sides by $n$ and using your two decompositions of $n$ gives $$ n=nxn_1+nyn_2=hn_2xn_1+kn_1yn_2\\=(hx+ky)n_1n_2 $$
Arthur
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$ a,b\mid n\iff ab\mid an,bn\iff ab\mid (an,bn)\!=\! \color{#c00}{(a,b)}n\,\ [= n\ \ {\rm if}\ \ \color{#c00}{(a,b)= 1}]$
Corollary $\,\ a,b\mid n \iff \dfrac{ab}{(a,b)}\mid\, n,\ $ thus $\,\ {\rm lcm}(a,b) = \dfrac{ab}{(a,b)}$
Bill Dubuque
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What's the difference between $ab \text{ } | \text{ } an,bn$ and $ab \text{ } | \text{ } (an,bn)$? – Oliver G Aug 08 '16 at 20:33
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1@OliverG $\ (x,y) ,$ is standard notation for $,\gcd(x,y)., $ That equivalence is the universal property of the gcd (see here), usually taken as the definition of the gcd in more general rings (which may have no notion of magnitude). – Bill Dubuque Aug 08 '16 at 20:40