Show that if $a$, $b$ are positive integers, then we have: $\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.
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Below are a few proofs, starting with two from my sci.math post on 2001/11/10
First, recalling that gcd distributes over lcm we obtain
$$(a\!+\!b,[a,b])\, =\, [(a\!+\!b,a),(a\!+\!b,b)]\, =\, [(b,a),(a,b)]\, =\, (a,b)$$
Or, cancelling $\,(a,b)\,$ homo reduces to case $\,(a,b) = 1\,$ so $\,ab = [a,b],\ $ so by Euclid's Lemma
$$\begin{eqnarray}(a\!+\!b,\color{#c00}a) = (b,a)= 1\\ (a\!+\!b,\color{#c00}b)=(a,b)=1\end{eqnarray}\ \Rightarrow\ 1 = (a\!+\!b,\color{#c00}{ab}) = (a\!+\!b,[a,b])$$
For variety here is yet another using $\rm\ (a,b)\ [a,b]\ =\ \color{#c00}{ab}\ \ $ and basic gcd laws:
$\rm\quad (a,b)\ (a\!+\!b,\, [a,b])\, =\, (aa\!+\!ab,\, ab\!+\!bb,\, \color{#c00}{ab})\ =\ (aa,\,bb,\,ab)\, =\, (a,b)^2$
By the way, recall that the key identity in the second proof arose the other day in our discussion of Stieltjes $\rm\ 4\:n+3\ $ generalization of Euclid's proof of infinitely many primes. Here's a slicker proof:
Lemma $\rm\ \ (a\!+\!b,\,ab) = 1 \iff (a,b) = 1$
Proof $\rm\ \ \ (\color{#90f}a,b)^2 \color{#C00}{\large\subseteq} (a\!+\!b,\,ab) \color{#0a0}{\large \subseteq} (a,b)\ \ $ since, $ $ e.g. $\rm\ \ \color{#90f}{a^2} = a(a\!+\!b)-ab\color{#c00}{\large \in} (a\!+\!b,\,ab)$.
$\rm\ 1\in (a\!+\!b,\, ab) \color{#0a0}{\large\subseteq}(a,b)\Rightarrow 1\in (a,b).\:$ Conversely $\rm\ 1 \in (a,b) \Rightarrow 1 \in (a,b)^2\color{#c00}{\large\color{#c00}\subseteq} (a\!+\!b,\,ab)$
Bill Dubuque
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Another Dubuquesque attempt; for legibility, write $d=\gcd(a,b)$: \begin{align*} \gcd\Bigl(d(a+b), ab\Bigr) &= \gcd\Bigl(d(a+b), ab, ab\Bigr)\\ &=\gcd\Bigl(d(a+b),\ ab-a(a+b),\ ab-b(a+b)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ a^2,\ b^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a^2,b^2)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a,b)^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ d^2\Bigr)\\ &= d\gcd\Bigl(a+b,d\Bigr)\\ &= d\gcd\Bigl(a+b,\gcd(a,b)\Bigr)\\ &= d\gcd(a,b)\\ &= \gcd(a,b)\gcd(a,b). \end{align*}
(Second line uses the fact that $a(a+b)$ and $b(a+b)$ are both multiples of $d(a+b)$).
Now divide through by $\gcd(a,b)$ to get the desired result.
Arturo Magidin
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Perhaps overkill, but if you accept the 'distribution law' $(x,[y,z])=[(x,y),(x,z)]$ stated at Wikipedia, then it is easy:
$(a+b,[a,b])=[(a+b,a),(a+b,b)]=[(b,a),(a,b)]=(a,b)$
where in the second equality I used the easy fact $(x,y)=(x,y \mod x)$.
wildildildlife
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Start by writing $a=d a'$, $b=d b'$, where $d=(a,b)$.
lhf
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Ok. Let d=(a,b), and l=[a,b]. Then d=(a+b,l). But a|l and b|l which means that l=am=bn for some positive integers m,n. Know also that a=da' and b=db' for some positive integers a', and b'. l=da'b' and (a',b')=1. Also ab=ld. – kira Feb 11 '11 at 17:03
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Let (a+b,l)=gcd(a+b,bn). Then gcd(a+b,bn)|a+b and gcd(a+b,bn)|bn by the definition of gcd. But gcd(a+b,bn)|(a+b)-b by the linear combination property. So, gcd(a+b,bn)|a. Similarly gcd(a+b,bn)|b.So, gcd(a+b,bn)|(a,b) and vice versa. So, (a,b)=(a+b,[a,b]). – kira Feb 11 '11 at 17:20
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@kira: Next time, write it as an addendum to your question; that way, the activity is reflected in the main page. I'll read it shortly and comment. – Arturo Magidin Feb 11 '11 at 21:08
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@Kira: You go from $\gcd(a+b,bn)|bn$ and $\gcd(a+b,bn)|a+b$ to $\gcd(a+b,bn)|(a+b)-b$. This does not follow from the linear combination property: you know $\gcd(a+b,bn)$ divides any linear combination of $a+b$ and $bn$, but this is a linear combination of $a+b$ and $b$, not $bn$. So you need to justify that assertion somehow, you haven't done so. – Arturo Magidin Feb 11 '11 at 21:10
sci.mathwith "SIMPLER...", I've finally started looking for arguments along these lines, even if they aren't the best one available just yet. – Arturo Magidin Feb 11 '11 at 22:04