Relation between $\gcd(a, b)$ and $\gcd(a+b, \operatorname{lcm}(a, b))$

Question

I've been trying to figure out a relation between $\gcd(a, b)$ and $\gcd(a+b, \operatorname{lcm}(a, b)).$ I know $$ \gcd(a, b) \mid (a+b). $$ And, since $a \mid \operatorname{lcm}(a, b)$ I have $$ \gcd(a, b) \mid \operatorname{lcm}(a, b). $$ Therefore, $$ \gcd(a, b) \leq \gcd(a + b, \operatorname{lcm}(a, b)). $$ I don't know the exact relation but a great deal of problems in my textbook are done as if they are equal. I shall be obliged if someone hints me as to how I shall proceed to establish the equality, if at all exists.

Answer 1

Let $d:=\gcd(a,b)$, $a=dm$, and $b=dn$, where $m$ and $n$ are integers (which are relatively prime). Then, $\text{lcm}(a,b)=dmn$. We want to prove that $$d=\gcd(a,b)=\gcd\big(a+b,\text{lcm}(a,b)\big)=\gcd\big(d(m+n),dmn\big)=d\,\gcd(m+n,mn)\,.$$ Thus, it suffices to show that $\gcd(m+n,mn)=1$. If this were not true, then there would exist a prime $p$ that divides both $m+n$ and $mn$, for which $p$ must divide at least one of $m$ and $n$, but then, as $p$ divides $m+n$, we conclude that $m$ and $n$ are both divisible by $p$, which is absurd, for $m$ and $n$ are relatively prime. The claim follows.

Answer 2

Suppose $p$ is a prime number and $p^r$ is the largest power of $p$ that divides $\operatorname{lcm}(a,b)$. Thus $p^r$ divides either $a$ or $b$.

Let $p^s$ be the largest power of $p$ that divides $a$ and $b$. Then $s\le r$.

Let $p^t$ be the largest power of $p$ that divides $a+b$. Clearly $s\le t$.

Now $p^{\min(r,t)}$ is the largest power of $p$ that divides $ \gcd(a + b, \operatorname{lcm}(a, b))$.

But $p^{\min(r,t)}$ divides $a+b$ and either $a$ or $b$, hence divide both, so $ \min(r,t) \le s$, so $s=\min(r,t)$.

But it is clear that $p^s$ is also the largest power of $p$ that divides $\gcd(a,b)$.

Since this is true for an arbitrary prime $p$, they are equal.