Sps that ab are integers with gcdab1 if not both a and b are

Sps that a,b are integers with gcd(a,b)=1

if not both a and b are odd , then

gcd ( a+b ,a-b)=1 prove on and give a comfertable example

Solution

Let a and b be relative prime integers, which are not both odd, so that gcd(a, b) = 1. Since a and b are not both odd, one of these is even and the other odd, for, otherewise, if both of these are even, then gcd( a, b) 2, which is not correct. Then a + b and a - b are both odd. Let us also assume that gcd( a +b, a – b) 1. Then there is an integer p > 1 such that a + b = p (a – b). Then p also has to be odd as a + b and a - b are both odd. Also, pa - a = pb + b   or a( p -1) = b( p+1) or a = [(p+1)/ (p-1)]b or, a/b = (p+1)/( p -1) . Now , since p is odd, both p + 1 and p -1 are even so that (p+1)/( p -1) 2 which means that gcd( a,b) 1. This is not true. Therefore gcd( a+b , a-b) =1. Let a = 5 and b = 2. Then a + b = 5 + 2 = 7 and a – b = 5 - 2 = 3. Now, we know that gcd( 5 , 2) = 1. Also, gcd ( 7, 3) = 1.