How to show that if $ad - bc = \pm 1$, then $(a+b, c+d) = 1$. I'm in a number theory course and I'm struggling to prove this effectively.
My idea was to use Bezout's Identity, but I can't see the application clearly.
Thanks for any help,
Neurax
 A: for integers $r,s$ if you can find $x,y\in \mathbb Z$ such that $xr+ys=\pm 1$ then $gcd(r,s)=1$. So:
$d(a+b)-b(c+d)=da+db-bd-bc=\pm 1$.
A: Simplify $d(a+b)-b(c+d)$.
Note that we do not need the "hard" direction of Bezout, only the easy direction.
A: Hint $\,\ \begin{eqnarray}\rm\ d\mid a\!+\!b\\ \rm d\mid c\!+\!d\end{eqnarray}\ \Rightarrow\: \rm mod\ d\!:\, $ $\begin{eqnarray}\rm b\equiv -a\\ \rm c\equiv -d\end{eqnarray}\rm \ \Rightarrow\:bc\equiv ad = bc\pm1\:\Rightarrow\:0\equiv \pm1,\ $ i.e. $\rm\ d\mid \pm1 $
Alternatively, $ $ let $\rm\ x,y = 1 = \Delta\ $ below $\:\Rightarrow$ $\rm\ gcd(a\!+\!b,c\!+\!d) = gcd(X,Y)\mid \Delta\, gcd(x,y) = 1.$
$\rm{\bf Lemma}\quad\ \ \begin{bmatrix} X \\\\ \rm Y\end{bmatrix}\ =\ \begin{bmatrix} a & \rm b \\\\ \rm c & \rm d \end{bmatrix}\ \begin{bmatrix} x \\\\ \rm y \end{bmatrix}  \ \ \ \Rightarrow\ \ \ gcd(X,Y)\mid \Delta\,gcd(x,y) ,\quad \Delta\ =\ ad\!-\!bc $
$\rm \begin{array}{}{\bf Proof} & \rm \ d\ X - b\, Y\ =\ \Delta\,x \\
\rm By\ Cramer\ \ \ \  & \\ &\rm\!\!\!  {-}c\ X + a\, Y\ =\ \Delta\,y \end{array}\quad so\quad \Bigg\lbrace\begin{eqnarray} d=gcd(X,Y)\mid X,Y\ \Rightarrow\ d\mid \Delta\:x,\:\Delta\:y\\ \\ \rm \Rightarrow\ d\ |\ gcd(\Delta\:x,\Delta\:y)\, =\, \Delta\ gcd(x,y)\end{eqnarray}$
