Vtyjkyuk

I've been trying to figure out a relation between $gcd(a, b)$ and $gcd(a+b, operatornamelcm(a, b)).$ I know
$$
gcd(a, b) | (a+b).
$$
And, since $a | operatornamelcm(a, b)$ I have
$$
gcd(a, b) | operatornamelcm(a, b).
$$
Therefore,
$$
gcd(a, b) leq gcd(a + b, operatornamelcm(a, b)).
$$
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.

Let $d:=gcd(a,b)$, $a=dm$, and $b=dn$, where $m$ and $n$ are integers (which are relatively prime). Then, $textlcm(a,b)=dmn$. We want to prove that $$d=gcd(a,b)=gcdbig(a+b,textlcm(a,b)big)=gcdbig(d(m+n),dmnbig)=d,gcd(m+n,mn),.$$
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.

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

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

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

Now $p^min(r,t)$ is the largest power of $p$ that divides $ gcd(a + b, operatornamelcm(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.