Question

I really need help. It's about number theory. Would yo be so kind help me?

Answer 1

Use the fact that if `d|n`, then `d|n^2` and if (`d|n` and `d|m`), then `d|n-m`

Explanation:

Suppose that `p=1` or `p` is prime, and that `p|a+b` and `p|a^2-ab+b^2`

From `p|a+b` , we conclude that `p|a^2+2ab+b^2`

Therefore, `p|3ab`, so `p|3` or `p|a` or `p|b`.

Now show that if `p` divides `a` pr `b`, then `p = 1`

Finish by showing that `9 cancel(|)(a+b,a^2-ab+b^2)`

Answer 2

See below.

Explanation:

Note that

`gcd(a+b,a^2-ab+b^2) = gcd((a+b)^2, a^2-a b + b^2)`

and also

`(a+b)^2 equiv (a²-a b + b^2) mod 3` because

`(a²-a b + b^2)-(a+b)^2 = 3 a b`

hence

`gcd(a+b,a^2-ab+b^2) = {(1),(3):}`