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

Mar 21, 2018

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 $p | {a}^{2} - a b + {b}^{2}$

From $p | a + b$ , we conclude that $p | {a}^{2} + 2 a b + {b}^{2}$

Therefore, $p | 3 a b$, 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 $9 \cancel{|} \left(a + b , {a}^{2} - a b + {b}^{2}\right)$

Mar 22, 2018

See below.

#### Explanation:

Note that

$\gcd \left(a + b , {a}^{2} - a b + {b}^{2}\right) = \gcd \left({\left(a + b\right)}^{2} , {a}^{2} - a b + {b}^{2}\right)$

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

$\gcd \left(a + b , {a}^{2} - a b + {b}^{2}\right) = \left\{\begin{matrix}1 \\ 3\end{matrix}\right.$