Question

If $G.C.D(a,b)=1$ then $G.C.D(a+b,a-b)$=?

• A. $1$ or $2$
• B. $a$ or $b$
• C. $a+b$ or $a-b$
• D. $4$

Answer 1

The correct option is A $1$ or $2$

It is given that GCD$(a,b)=1$

Let GCD$(a-b,a+b)=d$

$\Rightarrow d$ divides $a-b$ and $a+b$

there exists integers $m$ and $n$ such that

$a+b=m\times d$ ..........$\left(1\right)$

and $a-b=n\times d$ ..........$\left(2\right)$

Upon adding and subtracting equation $\left(1\right)$ and $\left(2\right)$ we get

$2a=(m+n)\times d$ ..........$\left(3\right)$

and $2b=(m-n)\times d$ ..........$\left(4\right)$

Since, GCD$(a,b)=1$(given)

$\therefore 2\times GCD(a,b)=2$

$\therefore GCD(2a,2b)=2$ since $GCD(ka,kb)=kGCD(a,b)$

Upon substituting value of $2a$ and $2b$ from equations $\left(3\right)$ and $\left(4\right)$ we get

$\therefore gcd((m+n)\times d,(m-n)\times d)=2$

$\therefore d\times gcd((m+n),(m-n))=2$

$\therefore d\times$ some integer$=2$

$\therefore d$ divides $2$

$\therefore d\le 2$ if $x$ divides $y,$ then $\left|x\right|\le \left|y\right|$

$\therefore d=1$ or $2$ since, gcd is always a positive integer.