NZMO Round One

Solution:
Let $d$ be a common divisor of both $(a + b)$ and $(a - b)$, and therefore divides linear combinations of $(a + b)$ and $(a - b)$. In particular,
$$d \mid [(a + b) + (a - b)] = 2a,\quad d \mid [(a + b) - (a - b)] = 2b.$$
As $a$ and $b$ don't share any common factors, the only common factors $2a$ and $2b$ have is $2$. Hence $d \mid 2$, which means that $d = 1$ or $2$. This means that the greatest common divisor cannot be larger than $2$.

We now show that both $1$ and $2$ can both occur.

Consider $a = 2$, $b = 1$
$$\gcd (2 + 1, 2 - 1) = 1.$$
Consider $a = 5$, $b = 3$
$$\gcd (5 + 3, 5 - 3) = 2.$$
Thus the only possible values for the greatest common divisor of $(a + b)$ and $(a - b)$ are $1$ and $2$.