# Thread: Prove if ad - bc = 1 or -1, then gcd(a+b, c+d) = 1

1. ## Prove if ad - bc = 1 or -1, then gcd(a+b, c+d) = 1

Problem:
Show that if ad - bc = 1 or -1, then gcd(a+b, c+d) = 1.

Work so far:
I know that since a(d) + b(-c) = 1 or -1, then the gcd(a,b) = 1. The same can be also said to show gcd(c,d) = 1, gcd(a,c) = 1, and gcd(b,d) = 1. I also know that gcd(a+b, c+d) divides a+b and c+d, so it will divide any linear combination of a+b and c+d.
I'm really at a loss of where to go from here.

2. Originally Posted by uberbandgeek6
Problem:
Show that if ad - bc = 1 or -1, then gcd(a+b, c+d) = 1.

Work so far:
I know that since a(d) + b(-c) = 1 or -1, then the gcd(a,b) = 1. The same can be also said to show gcd(c,d) = 1, gcd(a,c) = 1, and gcd(b,d) = 1. I also know that gcd(a+b, c+d) divides a+b and c+d, so it will divide any linear combination of a+b and c+d.
I'm really at a loss of where to go from here.

The equality $\displaystyle{ab+(-cd)=\pm 1$ means that $gcd(a,c)=gcd(a,d)=gcd(b,c)=gcd(b,d)=1$ .

Take it from here.

Tonio

3. It seems like you know that if you can write ax+by=1 for integers a,b, then gcd(x,y)=1. Can you express ad-bc as a linear combination of (a+b) and (c+d)?

4. It's enough to express $1$ as a linear combination of $a+b$ and $c+d$.

By inspection $(a+b)d+(-b)(c+d)=ad+bd-bc-bd=1$.
Then $(a+b)(-d)+(b)(c+d)=-1$.

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