Question

The hcf of(x,y)=1 then hcf of(x-y,x+y)=?

Answer 1

Answer
So the question asks to determine gcd(x+y,x−y)
given that gcd(x,y)=1.

write m=1a , n=1b

gcd(x+y,x−y)=gcd(1(a+b),1(a−b))=gcd(a+b,a−b)
divides their sum (which is
gcd(2a,2b)=2⋅gcd(a,b)=2

gcd(a+b,a−b)=2 if they are of the same parity (necessarily both a and b are odd) and gcd(a+b,a-b) =1 if a and b are of opposite parity. In fact, it equals 1 or 2).

To summarize with gcd(m,n)=1

gcd(m+n,m−n)=2gcd(m,n)=2
(if both a and b are odd)

gcd(m+n,m−n)=gcd(m,n)=1
(if a and b are of opposite parity)

Answer 2

Answer

So the question asks to determine gcd(x+y,x−y)

given that gcd(x,y)=1.

write m=1a , n=1b

gcd(x+y,x−y)=gcd(1(a+b),1(a−b))=gcd(a+b,a−b)

divides their sum (which is

gcd(2a,2b)=2⋅gcd(a,b)=2

gcd(a+b,a−b)=2 if they are of the same parity (necessarily both a and b are odd) and gcd(a+b,a-b) =1 if a and b are of opposite parity. In fact, it equals 1 or 2).

To summarize with gcd(m,n)=1

gcd(m+n,m−n)=2gcd(m,n)=2

(if both a and b are odd)

gcd(m+n,m−n)=gcd(m,n)=1

(if a and b are of opposite parity