gcd(odd, odd)

首先，２是一個Common Divisor，A + B 同 A - B都係雙數。

如果 GCD(A + B, A - B) > 2，那麼GCD(A + B, A - B) = 2K, K > 1

GCD(A + B, A - B) = 2K
GCD(A + B, A + B - (A - B)) = GCD(A + B, 2B) = 2K，亦即B能整除K。
GCD(A + B, A + B + (A - B)) = GCD(A + B, 2A) = 2K，亦即A能整除K。

GCD(A, B) >= K >1

那就跟命題矛盾了。

所以 GCD(A + B, A - B) = 2。

問非所答!!

Part I
Given that both A and B are odd numbers, it is easy to see that both (A+B) and (A-B) are even numbers. Therefore 2 is a common factor of (A+B) and (A-B).

Part II
Assuming that there is an integer M greater than 2 which is also a common factor of (A+B) and (A-B), then we have
(A+B) = aM ; and
(A-B) = bM where a and b are also integers.

Then we have 2A=(a+b)M and 2B = (a-b)M. It is easy to see that both (a+b) and (a-b) are also integers. Hence M is also a common factor of (2A) and (2B). However this contradicts with the assumption that the greatest common factor of A and B is 1 (which implies that the greatest common factor of (2A) and (2B) is 2).

By Part I and II, 2 is the greatest common factor of (A+B) and (A-B).

A = 15, B = 5
gcd( A+B , A-B)
= gcd(20,10)
= 10

題目自己作架? 咪亂黎啦, 好無聊