452. Natural numbers $a$ and $b$, where $a>b$, are coprime. Find all values of the greatest common divisor of the numbers $a+b$ and $a-b$.
$\triangle$ Let

$$
\text { GCD }(a+b, a-b)=d
$$

Then

$$
(a+b): d, \quad(a-b): d
$$

Therefore, the sum and difference of the numbers $a+b$ and $a-b$ are divisible by $d$. The sum of these numbers is $2 a$, and the difference is $-2 b$. We get:

$$
2 a: d, \quad 2 b: d .
$$

But the numbers $a$ and $b$ are coprime by condition, so 2 is divisible by $d: 2: d$. Hence $d=1$ or $d=2$.

Are both of these cases possible? Both: $d=1$, if the numbers $a$ and $b$ have different parity, and $d=2$, if they are odd.

Answer: 1 or 2.