If c is a common factor of two different numbers, that means the numbers can be written as a*c and b*c, and their difference is |a-b|*c. Therefore, the closest they can be (assuming a != b) is when |a-b|=1, making the difference 1*c = c. In other words, if GCF(a,b) = c, |a-b| >= c. So... Since 2 divides all even numbers, and no odd numbers, If a is even, a+2 is also even, and GCF(a,a+2) = 2. If a is odd, GCF(a,a+2) = 1.

Alternatively, you could find the answer with one iteration of the euclidean algorithm: where a > b GCD(a,b) = 1) GCD(b,c) where c is the remainder to a/b; OR 2) b if the remainder is 0. So: if a>2: GCD(a,a+2) (a+2)/a = 1r2 GCD(a,a+2) = GCD(a+2,2) = 1 if a+2 is odd, 2 if a+2 is even. if a=2: GCD(a,a+2) (a+2)/a = 4/2 = 2r0 GCD(2,4) = 2 if a=1: GCD(a,a+2) (a+2)/a = 3/1 = 3r0 GCD(1,3) = 1