Show that if $a$ and $b$ are positive integers then $(a, b)=(a + b, [a, b])$.
I was thinking that since $[a, b]=LCM(a, b)=\frac{ab}{(a, b)}$ that if $d= (a + b, [a, b])$, then $d|[a,b]$ and thus $d|(a, b)$ since $(a, b)|[a, b]$
Then I would just have to prove that $(a+b, (a, b)) = (a, b)$.
Is that the right way to go?
Another way : Let $(a,b)=d$ and $\displaystyle \frac aA=\frac bB=d\implies (A,B)=1$
So, $\displaystyle(a+b, [a,b])=(d(A+B), dAB)=d(A+B,AB)$
Now, if $D$ divides both $ A+B, AB; D$ will divide $A(A+B)-AB=A^2$ and $D$ will divide $B(A+B)-AB=B^2$
$\displaystyle\implies D$ will divide $(A^2,B^2)=(A,B)^2=1\implies D=1$
I assume the direction $(a, b) \mid (a+b, [a,b])$ poses no problem.
For the reverse divisibility, set $d = (a, b)$ and let $x, y \in \mathbb Z$ with $ax+ by = d$. Then for all $n \in \mathbb Z$, $$(a+b)\left( x + n\frac{b}d\right) + b \left(y-x - n\frac{a+b}d\right) = d$$ We choose $n$ such that the second term in the LHS is divisible by $a$, and we will have a linear combination of $a+b$ and $[a,b]$ that equals $d$. Take $n = y(y-x)$. We have $$1 - y\frac{a+b}d = \frac ad (x-y)$$ Multiplying both sides by $y-x$ and plugging this in gives $$(a+b)\left( x + y(y-x)\frac{b}d\right) + b \frac ad (x-y)(y-x) = d$$ as desired.
