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The generalized version of Euclid's lemma states that if $k|mn$ and that $\gcd(k, m) = 1$ then $k|n$. However, I noticed an alternative way of proving questions such as: if $2|n$ and $3|n$ show $6|n$ by getting $n = 3p = 2q$ for some integers $p, q$ and considering $3n - 2n = n = 3(2q) - 2(3p) = 6(p - q)$ and as $p - q$ is an integer we are done. This works for larger problems such as getting $42|n$ from $2|n$, $3|n$ and $7|n$ by considering $21n - 14n - 6n = n = 21(2p) - 14(3q) - 6(7r) = 42(p - q - r)$.

What is this technique called? Does it always work? Can you point me in the right direction?

James47
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dakenSoren
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    It is one way of proving that Euclid's lemma. If $(k,m)=1$ then there are $a,b\in\mathbb{Z}$ such that $ka+bm=1$. Then $n=(ka+bm)n=kan+mnb$ which is multiple of $k$, because $mn$ is a multiple of $k$. So, yes, you can always replace the use of that Euclid's lemma by the use of some integer combination that sums $1$. It is like replacing the lemma by its proof everytime you use it. – Pp.. Jan 20 '15 at 18:53
  • @Pp.. That makes sense, thank you. – dakenSoren Jan 20 '15 at 19:01
  • See this answer. – Bill Dubuque Jan 20 '15 at 19:58

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Of course. The way you might prove Euclid's Lemma - or your related fact - is by showing the equation $ax+by=1$ has integer solutions when $\gcd(a,b)=1$ (try it!), and you are essentially following that proof.

djechlin
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