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Notation: $(a,b)$ is the gcd of $a$ and $b$, and $[a,b]$ is their lcm.

Problem: If $a,b\in\Bbb Z_+$, show that $(a+b,[a,b])=(a,b)$

My solution: Let $d'=(a,b)$ and $d=(a+b,[a,b])$. Clearly $d'$ divides $a+b$ and $[a,b]$, so $d'\mid d$.

Let's show that $d\mid a$. We know (this is shown in a previous problem) that $ab=[a,b]d'$. If we let $a=a'd'$ and $b=b'd'$ we can write $a'b=ab'=[a,b]$. It's shown also in other previous problem thet $(a',b')=1$, so there exist integers $x,y$ such that $a'x+b'y=1$.
Since $d\mid a+b$, $d\mid aa'+a'b$. Therefore $d\mid aa'$. And then $d\mid aa'x+ab'y=a$.
In a similar way, we can show that $d\mid b$. Thus, $d\mid d'$ and we are done.

Question: In the context where the problem is, I find this solution too complex, and I'm afraid that I'm missing something. Is there a simpler way to show that $d\mid d'$?

Reference: This exercise is in Ireland & Rosen's A Classical Introduction to Modern Number Theory, 2nd edition, p. 15.

Bill Dubuque
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ajotatxe
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    If you write $a = d'\cdot \alpha,; b = d'\cdot \beta$, you have to show that $\gcd(\alpha+\beta, \alpha\cdot \beta) = 1$. – Daniel Fischer Aug 31 '17 at 19:40
  • So, for any prime $p$ dividing $\alpha\beta$, if $p\mid \alpha$ then $p$ does not divide $\beta$, so it does not divide $\alpha+\beta$ either, and similarly if $p\mid \beta$. – ajotatxe Aug 31 '17 at 19:49

2 Answers2

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Simpler: by gcd * lcm law $\,(a,b)[a,b] = \color{#c00}{ab}\,$ and gcd laws (associative, commutative, distributive)

$\ (a,b)\ (a\!+\!b, [a,b]) = (aa\!+\!ab,\,ab\!+\!bb,\, \color{#c00}{ab}) = (aa,bb,ab) = (a,b)^2,\,$ then cancel $\,(a,b).$

Alternatively we can cancel $(a,b)$ reducing to the special case $\,(a,b)=1\,\Rightarrow\,(a\!+\!b,ab) = 1,\,$ which follows by Euclid using $\,(a\!+\!b,a) = (a,b)[=1] = (a\!+\!b,b).\,$

Another very simple way is to use that gcd distributes over lcm. See this answer for these proofs.

Bill Dubuque
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Label $(a,b) = d$ and define $a', b'$ so that $a = a'd; b=b'd$.

We know as fundamental properties that $(a',b') = 1$ and that $[a,b]=a'b'd$. (If we don't know these it is very easy to prove).

We also have a fundamental property that $(cg,eg) = g*(c,e)$. (Also easy to prove.)

So $(a+b, [a,b]) = (d(a' + b'), da'b')= d(a'+b', a'b')$.

So it suffices to show given relatively prime $a',b'$ that $a'b'$ and $a' + b'$ are relatively prime.

That is easy. Any prime factor $p$ of $a'b'$ will be a prime factor of either $a'$ or $b'$ but not the other. So it will not be a factor of $a' + b'$. So $a'+b'$ and $a'b'$ can not share any prime factors.

That's it.

fleablood
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  • Thiat's the 2nd method I mention, except you conclude using the prime divisor property vs., Euclid's Lemma. – Bill Dubuque Aug 31 '17 at 22:48
  • You hadn't posted when I had started my posting. – fleablood Aug 31 '17 at 22:51
  • Btw, we don't need to use the full gcd * lcm law $,(a,b)[a,b] = ab,$ but only the special case $,(a,b) = 1,\Rightarrow, [a,b] = ab,$ after we cancel $,d = (a,b),$ throughout the original equation. Then it reduces to $,1 = (a'+b',[a',b']) = (a'+b',a'b')$ in your notation. Normally this is said more concisely, e.g. wlog $(a,b) = 1.$ since by distributivity of gcd and lcm we can cancel $,d=(a,b).,$ The first comment is to help readers understand the relationship between the methods - nothing more. – Bill Dubuque Sep 01 '17 at 20:55