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For a proof in my textbook it is asserted that if $d=(n,a), n=db, a=dc$ for suitable $b,c\in\mathbb{Z}$ then $(b,c)=1$.

I am not very good with number theory and this boggles me a bit. I am sure it is not so difficult to realize but I am stumped right now. Could someone please help me out?

An0ku
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1 Answers1

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We start with

  • $d = (n,a)$. That is, $d$ is the greatest common divisor of $n$ and $a$. Any other common divisor of $n$ and $a$ divides $d$.
  • Then we factor $n$ and $a$: $n = db$ and $a = dc$ for some $b \in \Bbb{Z}$ and some $c \in \Bbb{Z}$. We already know $d$ is a divisor of $a$ and $n$, so we can always find the quotients $n/d = b$ and $a/d = c$.
  • Finally we look at the common divisors of $b$ and $c$. There shouldn't be any -- $b$ and $c$ are what was left after we extracted the greatest common divisor from $a$ and $n$.

Notice that if $g > 1$ divides both $b$ and $c$, then we can write $b = gh$ and $c = gk$ for suitable integers $h$ and $k$. Then $n = d b = d g h$ and $a = d c = d g k$. These have the evident common divisor $dg$. Since $g > 1$, we must have $dg > d$. But $d$ is the greatest common divisor of $a$ and $n$, contradicting that $dg$ is an even greater common divisor. Therefore, $g \not > 1$. Therefore, $g = 1$ : that is, the greatest common divisor of $b$ and $c$ (which have already had their common divisors factored away) is $1$.

Eric Towers
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