Raghul101
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)
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).
Answers & Comments
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
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