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I'm having trouble understanding the indicated line from the solutions

Do a proof by contradiction by assuming there exists an $a > 1$ such that $a \mid \frac{n}{d}$ and $a \mid \frac{m}{d}$.

This implies (I don't understan why!) $ad\mid n$ and $ad\mid m$,

but $ad>d = \gcd(m,n)$, a contradiction.

Michael Hardy
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saner
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    Since $a|\frac{n}{d}$ and $a|\frac{m}{d}$, $ak=\frac{n}{d}$ and $al=\frac{m}{d}$ ($k,l$ are integers). This suggests $adk=n$ and $adl=m$ or $ad|n$ and $ad|m$. Thus $ad\neq d$ is the gcd which contradicts with the assumption that $d$ is the gcd. – justadzr Dec 12 '19 at 00:59
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    If we have an integer $b$ such that $ab=\frac nd$ then $abd=n$. – lulu Dec 12 '19 at 00:59
  • No need to use contradiction. Let $a$ be a common divisor of $\frac{n}{d}$ and $\frac{m}{d}$. Then $ad$ is a common divisor of $n$ and $m$ and so divides $\gcd(n,m)=d$. Therefore, $a=1$. – lhf Dec 12 '19 at 01:13
  • @YourongZang, amazing, I understood the proof, but I wouldn't know how to start one on my own! – saner Dec 12 '19 at 02:11
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    $\ a\mid b,\Rightarrow, ad\mid bd.\ $ OP is case $,b = n/d,$ so $,bd = n\ \ $ – Bill Dubuque Dec 12 '19 at 02:30
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    @eriksan Next time you can try to start from the definitions and try to get something from the definitions. – justadzr Dec 12 '19 at 03:05
  • $a\mid \frac nd$ means that $\frac nd = ak$ for some $k$ and $a\mid \frac md$ means that \frac nd = aj$ for some $j$. So $n = d\cdot \frac nd= d(ak) = (ad)k$ and $m = d\cdot\frac md=d\cdot (aj) = (ad)j$. So $ad|n$ and $ad|m$. – fleablood Dec 12 '19 at 06:12

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