GCD of $(a+b,a^2-ab+b^2)= 1$ or $3$

Question

I have been given that $(a,b)=1$. I need to prove that $(a+b,a^2-ab+b^2)=1$ or $3$. I am not able to reduce the expressions $a+b$ and $a^2-ab+b^2$ in a way that I could reach the proof of the given proposition. Any help would be appreciated.

Answer 1

Let $d=(a+b,a^2-ab+b^2)$.

Since $d|(a+b) $ and $d|(a^2-ab+b^2)$ $\implies d|[(a+b)^2-a^2+ab-b^2]$ $\implies d|3ab$.

Therefore, $d|[3b(a+b)-3ab]$, i.e., $d|3b^2$.

Similarly, it can be shown that $d|3a^2$.

Therefore, $d|(3a^2,3b^2) \implies d|3(a,b)^2=3$.

Hence, $(a+b,a^2-ab+b^2)=1$ or $3$

Answer 2

$a^2 - ab + b^2 = (a+b)^2 - 3ba$

so $\gcd (a+ b, a^2 -ab + b^2) = \gcd (a+b, 3ab)$

Now $\gcd(a+b,a) = \gcd(b,a) = 1$ and $\gcd(a+b, b) = 1$ so

$\gcd(a+b, 3ab) = \gcd(a+b, 3) = \{1|3\}$.

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P.S. We have know that i) $\gcd(m,n) = \gcd(m, n \pm km)$ which is a basic result

ii) If $\gcd(m,d) =1$ then $\gcd(m, nd) = \gcd(m,n)$ which is almost as basic.

Answer 3

If a prime $p$ divides $a+b$ then it also divides $a^2+2ab+b^2$. But $p$ does not divide $a$ nor $b$, because $a$ and $b$ are coprime.

Moreover, if $p$ divides $a^2-ab+b^2$ then $p$ divides $a^2+2ab+b^2-(a^2-ab+b^2)=3ab$. Then $p= 3$.

So $\gcd(a+b,a^2-ab+b^2)=3^t$. Let's show that $t\le 1$.

If $9$ divides $a+b$, or, equivalently, $a\equiv -b\pmod 9$, then $$a^2-ab+b^2\equiv 3a^2\equiv \pm3\pmod 9$$

Thus $9$ does not divide $\gcd(a+b,a^2-ab+b^2)$. This completes the proof.