Understanding proof that involves gcd and linear combinations

Question

The problem reads like this:

Prove that if $gcd(a, b) = 1$, then $gcd(a + b, a - b)$ equals to 1 or 2.

Solution: From the linear combinations

$(a + b) * 1 + (a - b) * 1 = 2a$

$(a + b) * 1 + (a - b) * (-1) = 2b$

we know that $gcd(a + b, a - b)$ divides both $2a$ and $2b$. Since $gcd(a, b) = 1$, we conclude that $gcd(a + b, a - b)$ divides 2. Consequently $gcd(a + b, a - b)$ is either 1 or 2.

I tried to follow this proof, but I really got stuck at the From the linear combinations ... we know that ... part. Where do these linear combinations come from? Especially the $(a + b) * 1 + (a - b) * (-1) = 2b$ part is hard to understand, where is the $-1$ coming from?

I'm familiar with Euclids algorithm, but don't understand how the linear combinations above fit into the picture.

Answer 1

Let $d=\gcd(a+b,a-b)$. This means that $d\mid (a+b)$ and $d\mid(a-b)$ and $d$ is the largest such integer. So $d$ must divide any linear combination of $(a+b)$ and $(a-b)$. Now we know that $\gcd(a,b)=1$. So we will make use of this fact. So we take the given linear combinations OR in fact anything else that will work. Then by those linear combinations we see that $d\mid2a$ and $d\mid2b$ because $d$ divides any linear combination of $(a+b)$ and $(a-b)$. Since $\gcd(a,b)=1$, we have integers $r,s$ such that $ar+bs=1$, whence $(2a)r+(2b)s=2$. But since $d\mid 2a$ and $d\mid 2b$ we have $d\mid ((2a)r+(2b)s)$. So $d\mid 2$ and you have $d=1$ or $2$.

Answer 2

For a pair of integers $n,m$ we have that $gcd(n,m)$ divides any linear combination of $n$ and $m$. I.e. we have $$ gcd(n,m) | kn+jm \qquad \forall k,j \in \mathbb{Z}. $$ By choosing $k=1, j=-1$ we get the above linear combination.

The linear combinations in the proof are just chosen to get statements that help your conclusions, they are not derived anywhere from in this proof.

Answer 3

Let gcd$(a+b,a-b)=g$. Clearly $g$ divides $(a+b)$ as well as $(a-b)$.
So if $(a+b)=kg$ and $(a-b)=lg$ then, $2a=g(l+k)$ and $2b=g(k-l)$. Equivalently $g$ divides $2a$ and $2b$ and hence the gcd$(2a,2b)=2$.
Therefore g is equal to 1 or 2.