Common divisor of $a+b$ and $ab$. If $\gcd(a,b) =1$.
Why does $\gcd(a+b,ab)=1$ ?
I know that if $\gcd(a,b)=1$ then there exists $u$ and $v$ where $au+bv=1$. But I can't seem to relate it to $a+b$ and $ab$.
 A: If $(a,b)=1$, Bezout's Identity says there are $x,y$ so that $ax+by=1$. Then
$$
(a^2x+b^2y)(\color{#C00000}{x+y})-(a-b)^2\color{#C00000}{xy}=(ax+by)^2=1
$$
Therefore, $(x+y,xy)=1$.
A: Hint: $a(a+b) - ab = a^2$ and $b(a+b) - ab = b^2$ so that $\mathrm{gcd}(a+b, ab)$ divides $a^2$ and $b^2$.
A: Since
$ab-a(a+b)
=-a^2
$
and
$ab-b(a+b)
=-b^2
$,
if
$d$ divides both
$ab$ and $a+b$,
then $d$ divides both
$a^2$ and $b^2$.
Since
$(a, b) = 1$,
$(a^2, b^2) = 1$,
so $d = 1$.
A: Here is a direct proof for those who enjoy Bezout fiddling.
$\quad \begin{align}
\gcd(a,b)=1\ \Rightarrow\ 1 &=\,  j a + k b\quad\text{for some }\ j,k\in\Bbb Z\\
&=\, (j\!-\!k)a + k(a+b)\\
&=\, (k\!-\!j)b  + j(a+b),\ \ \ \text{so multiplying this with above}\\
\Rightarrow\ 1 &=\quad\, m\,ab + n (a+b)\ \ \ \text{for some }\ m,n\in\Bbb Z \\
\Rightarrow\ 1 &=\ \gcd (ab,\ a+b)
\end{align}$
Remark $\ $ Using gcd laws eliminates the obfuscatory Bezout coefficients
$$ (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\ \ {\rm if}\ \ (a,b,c) = 1\qquad$$
Yours is the special case $\ c = a+b,\ $ and $\ (a,b)=1\ \,(\Rightarrow\, (a,b,c)=1)$
A: Hint $\ $ By here: $\ (c,a)=1=(c,b)\,\Rightarrow\, (c,ab)=1.\ $  Let $\ c = a+b.$
Remark $\ $ More generally $\ (a\!+\!b,\, {\rm lcm}(a,b))\, =\, (a,b).\,$ See here for a few proofs.
