If $\gcd (a,b) = 1$, what can be said about $\gcd (a+b,a-b)$?

Question

Suppose $a, b \in \mathbb{Z}$, $a > b$, and $\gcd (a,b) = 1$.

What can be said about $\gcd (a+b,a-b)$? Is it true in general that $\gcd (a+b,a-b) \leq 2$?

Answer 1

Hint If $d= \gcd(a+b,a-b)$ the $d| (a+b)+(a-b)$ and $d \mid (a+b)-(a-b)$.

Answer 2

You can see that

$$\gcd(a+b,a-b)=\gcd(a+b, a+b+a-b)=\gcd(a+b,2a)$$

but then this divides $2a$, and since any divisor of $a$ does not divide $b$, it must be that it divides $2$, i.e. it is either $1$ or $2$.

Answer 3

Hint $\ $ The Theorem below yields $\rm\ \gcd(a,b)\mid gcd(a\!-\!b,a\!+\!b)\mid 2\gcd(a,b)$

Theorem $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y),\ \ \Delta = \det A$

Proof $\ $ Inverting the linear map $\rm\,A\,$ by Cramer's Rule (multiplying by the adjugate) yields

$$\begin{eqnarray}\rm a\ x\, +\, b\ y &=&\rm X\\ \\ \rm c\ x\, +\, d\ y &\ =\ &\rm Y\end{eqnarray} \quad\Rightarrow\quad \begin{array}{}\rm \Delta\ x\ \ \ =\ \ \ \rm d\ X\, -\, b\ Y \\\\ \rm\Delta\ y\ =\ \rm -c\ X\, +\, a\ Y \end{array}\rm \ ,\quad\ \Delta\ =\ ad-bc\qquad $$

So, by the RHS system, $\rm\,\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\:\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)\ =\ \Delta\ gcd(x,y)\:.$
In particular $\rm\ n = \gcd(X,Y) \mid \Delta\, \gcd(x,y).\ $

And, by the LHS system, $\rm\ n\mid x,y\ \Rightarrow\ n\mid X,Y\ \Rightarrow\ n\mid\gcd(X,Y).$
In particular $\rm\ n = gcd(x,y)\mid \gcd(X,Y).\ \ \ $ QED