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I'm really stuck with this.

Let $a,b\in \mathbb{Z}^+$ such that $$ GCD(a,b) + LCM(a,b) = a+b $$ then either $a|b$ or $b|a$.

I tried using the fact that the GCD is a linear combination of the numbers or the equality $GCD \times LCM = ab $ without absolute value because $ab>0$.

Edit:

Apparently, when the sum of the GCD and LCM is the sum of $a$ and $b$ then one of them is the GCD.

Bill Dubuque
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Sergio Ivan
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2 Answers2

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Notation: $GCD(a,b)=g$, and $LCM(a,b)=l$.

It is well-known that $gl=ab$.

Substituting in your relation, we have, $g+\frac{ab}{g}=a+b$, which means: $g^2-(a+b)g+ab=0$.

So that, $(g-a)(g-b)=0$.

So $g=a$ or $g=b$, which means $a\mid b$ or $b\mid a$. QED.

amWhy
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Ritam_Dasgupta
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With no loss of generality we assume $a\leq b$. We are given the condition $(,)+(,)=+$ we seek to verify $a|b$.

$a$ divides the lcm by definition, if in addition $a|gcd$ then $a$ also divides $b=gcd + lcm -a$, which would suffice for the proof.

Does $a|gcd$? Since $gcd\times lcm = ab$ then $b=\frac{gcd\times lcm}{a}$ and substitute this into $(,)+(,)=+$ we have that $$(,)+(,)=+\frac{(,)+(,)}{a}$$ you find that $gcd / a$ is an integer, so a divides the gcd. Your claim is verified.

ago752
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  • This question seems not to meet the standards for the site. Instead of answering it, why not look for a good duplicate target, or help the user by posting comments suggesting improvements? Please also read the meta announcement regarding quality standards. – amWhy May 14 '21 at 20:53
  • You made a mistake substituting $b$. The numerator is $GCD(a,b) LCM(a,b)$, not $GCD(a,b) + LCM(a,b)$. So you can't conclude from the equality that $gcd/a$ is an integer. – jjagmath May 14 '21 at 21:52
  • By the way, something that stands out is that you never used the assumption that $a \le b$, so your proof couldn't be right. – jjagmath May 14 '21 at 21:54