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Let $c \in \mathbb Z$ and $k=LCM(a,b)$.

Show that for every $a,b \in \mathbb Z$ \ {$0$} if $a\mid c$ and $b\mid c$ then also $k\mid c$.

Please dont give me the solution just a tip. My thoughts were that $c \ge LCM(a,b)$ so $c$ is a multiple of $k$ $\Rightarrow$ $c=LCM(a,b)*m$ with $m \in \mathbb N$. Then I have to look at to cases:

Case 1:

$c=LCM(a,b)$

then $a$ divides $c$, $b$ divides $c$ and then also $k$ divides $c$.

Case 2:

$c\gt LCM(a,b)$

a divides c and b divides c $\Rightarrow ax=c=by $ with $x,y \in \mathbb Z$ and a,b divides $LCM(a,b)$ then a,b divides also the multiple of $LCM(a,b)*ax*by*n$ with $n \in \mathbb N$.

Im pretty sure at Case 1 but not at Case 2. Does it make sense? please give me rate and a tip. Thank you!

RobPratt
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Arji
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2 Answers2

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The set $\,S\,$ of all positive common multiples of $\,a,b\,$ is a nonsempty set of postive integers closed under positive subtraction, i.e. $\,n,m\in S, n>m\,\Rightarrow\, n-m\in S.\,$ Show that every element $\,s\,$ of such a set is a multiple of its least element $\,\ell = \min S\,$ (hint: $ $ if not then we would have $\,0\ne s\bmod \ell = s - \ell -\cdots -\ell\,$ lies in $\,S\,$ and is smaller than $\,\ell,\,$ contra minimality of $\,\ell)$

Bill Dubuque
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One idea is to look at prime factorisation of $a,b, \text{lcm} (a,b)$. Denote $\nu_p(x)=d$ if $p^d \mid x$ and $p^{d+1} \nmid x$. Now, according to the definition of least common multiple, we find that $\nu_p(c)=\max \{ \nu_p(a),\nu_p(b) \}$. We also have $\nu_p(k) \le \max \{ \nu_p(a),\nu_p(b) \}$. Thus, for all prime $p$ then $\nu_p(k) \le \nu_p(c)$ or $k \mid c$.

Tengu
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