Prove that $\gcd(a, b) = \gcd(a + b, \operatorname{lcm}(a, b))$

Question

Is it true that $$ \gcd(a, b) = \gcd(a + b, \operatorname{lcm}(a, b)) $$ for all integers $a$ and $b$? How to prove it?

For example, take $a = 48$ and $b = 60$. Now $\gcd(a, b) = 12$, $\operatorname{lcm}(a, b) = 240$, and $a + b = 108$. Now, the $\gcd(240, 108) = 12$. Does the relation hold in general for any two integers?

What is $\operatorname{HCF}$? What makes you think it satisfies this identity? What are some examples of this? What have you tried to prove it? This question needs context. Please see this page about how to ask a good question on this site. — Sammy Black, Jul 07 '22 at 02:42
upvote for curiosity, needs some definitions and examples added to this question. now i am wondering too. — jimjim, Jul 07 '22 at 02:48
For clarity, the question is asking whether $\gcd(a, b) = \gcd(a + b, \operatorname{lcm}(a, b))$ for all $a, b \in \mathbb{Z}$. — L. F., Jul 07 '22 at 03:14
HCF ("highest common factor") is a alternative term for greatest common divisor. — lonza leggiera, Jul 07 '22 at 03:15
Does this answer your question? $\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$ - found using an Approach0 search. — John Omielan, Jul 07 '22 at 03:48
Thanks for the answer @L.F. Yes John Omielan, it was my question. HCF is another name used for GCD. — Heerack Ganguly, Jul 07 '22 at 03:49
FYI, there's also quite a few other duplicates, e.g., Let $a,b\in \mathbb{N^*}$. Prove that $\gcd(a+b,\operatorname{lcm}[a,b])=\gcd(a,b)$. Is my proof correct?, Simpler solution for $,\gcd(a+b,{\rm lcm}(a,b)) = \gcd(a,b)$, Proving or disproving gcd and lcm, Prove that $\gcd(a,b) = \gcd (a+b, \gcd(a,b))$, Derive a simpler expression for $\gcd(a+b,\text{lcm}(a,b))$?, Relation between $\gcd(a, b)$ and $\gcd(a+b, \operatorname{lcm}(a, b))$, etc. — John Omielan, Jul 07 '22 at 03:54
Another interesting property of positive integers $a,b$ is $\gcd(a,b)\cdot $ lcm$(a,b)=ab.$ — DanielWainfleet, Jul 07 '22 at 05:28

L. F. · Answer 1 · 2022-07-07T06:19:27.650

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$$ \DeclareMathOperator{\lcm}{lcm} \newcommand{\ZZ}{\mathbb{Z}} $$

Given $a, b \in \ZZ^*$, we are trying to show that $\gcd(a, b) = \gcd(a + b, \lcm(a, b))$. (I will ignore the case where $a = 0$ or $b = 0$, in which case whether the statement holds depends on how you define $\gcd(0, n)$.)

Denote $g = \gcd(a, b) \in \ZZ^+$ and let $a = gx$ and $b = gy$ with $x, y \in \ZZ^*$ and $\gcd(x, y) = 1$. Now $\lcm(a, b) = gxy$, so

$$ \gcd(a + b, \lcm(a, b)) = \gcd(g(x + y), gxy) = g \cdot \gcd(x + y, xy). $$

Assume for the sake of contradiction that $d$ is some prime factor of $\gcd(x + y, xy)$. Then, $d$ divides $xy$. Since $\gcd(x, y) = 1$, $d$ divides exactly one of $x$ and $y$. But $d$ also divides $x + y$, so $d$ divides neither or both of $x$ and $y$, reaching a contradiction. Therefore, $\gcd(x + y, xy) = 1$, and $$ \gcd(a + b, \lcm(a, b)) = g. $$

edited Jul 07 '22 at 06:19

answered Jul 07 '22 at 03:26

L. F.

1,940

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Don’t you need $d$ to be prone to say that $d|xy$ implies that $d|x$ or $d|y$? – G Tony Jacobs Jul 07 '22 at 05:38
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@GTonyJacobs Indeed - thanks for pointing out the mistake! – L. F. Jul 07 '22 at 06:20
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Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 07 '22 at 06:47
@BillDubuque Thanks for the notice. I will keep that in mind for the future. – L. F. Jul 07 '22 at 14:35
Glad to hear that. – Bill Dubuque Jul 07 '22 at 14:45

Prove that $\gcd(a, b) = \gcd(a + b, \operatorname{lcm}(a, b))$

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