TITLE: Prove that $\gcd( a + a', b + b' ) = 1$ if $ab - a'b' = \pm 1$
QUESTION [5 upvotes]: Prove that $\gcd(a + a', b + b') = 1$ if $ab - a'b' = \pm 1$

My attempt was:
Case 1:
$ab - a'b' = 1 \implies \gcd(a, b') = 1$ and $\gcd(a', b) = 1$
Then is it sufficient to conclude that $\gcd(a + a', b + b') = 1$? 
Furthermore, when we write $\pm 1$, does it mean or or and?  
Thanks,

REPLY [10 votes]: Here's one way of looking at these problems. In general, integers $m$ and $n$ are relatively prime if and only if there are integers $x$ and $y$ such that $mx - ny = \pm 1$. You can write this in matrix form as 
$$det\pmatrix {m&n \cr y&x \cr} = \pm 1$$
Correspondingly, the condition you are given is that
$$det\pmatrix {a&b' \cr a'&b \cr} = \pm 1$$
Adding one row to another doesn't change the determinant, so you also have
$$det\pmatrix {a + a'&b + b' \cr a'&b \cr} = \pm 1$$
This gives that $a + a'$ and $b + b'$ are relatively prime.