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I need help to prove this: If $\gcd(a,b)=1$ then $\gcd(a+b, a^2-ab+b^2)$ is equal to $1$ or $3$. I have done this:

Let $d$ be the g.c.d. of $(a+b, a^2-ab+b^2)$, then $d$ divides $a+b$ and $d$ divides $a^2-ab+b^2$. That implies $d$ divides $(a+b)n + (a^2-ab+b^2)m$, for some $n,m$ integers. Let $n$ and $m$ be $a$ and $1$ respectively, then $d$ divides $2a^2+ b$. With the same argument but with $n=b$ we get $d$ divides $a^2+2b$. Then $d$ divides $3a^2+3b^2$. That implies $3a^2+3b^2 \geq d$, and we get that $3 \geq d$, because $\gcd(a^2+b^2)=1$. So $d$ must be $3$ or $1$ because if $d =2$, $d$ has to divide $a^2+2b$ and $2a^2+ b$, but we see that no. So $d=3$ or $1$. I don't know if I did it well.

Bill Dubuque
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SERGIO GARCÍA GARCÍA
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2 Answers2

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You started right. Let $d =\gcd(a+b, a^2-ab+b^2)$, then $d$ divides $a+b$ and $a^2-ab+b^2$. But from here it is pretty messy.

We can say $$d\mid (a+b)^2- (a^2-ab+b^2) = 3ab$$

So if prime $p\mid d$ we have $p\mid 3$ or $p\mid a$ or $p\mid b$. Say $p\nmid 3$ and say $p\mid a$. Since $p\mid a+b$ we have then $p\mid b$, a contradiction. So $p\mid 3$ and thus $d=3^n$.

Say $n\geq 2$, then $9\mid 3ab$ so $3\mid ab$ so $3\mid a$ or $3\mid b$. In both cases we have that $3$ divides then also the other number since $3\mid a+b$, a contradiciton. So $n\leq 1$ and we are done.

nonuser
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Well try to factor $a^2 -ab +b^2$ into terms of $a+b$ or $a,b$.

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

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

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

$\gcd(a+b, -3ab)=\gcd(a+b,3ab)$.

Now... $\gcd(a,b) =1$ so any factor of $a$ will not be a factor of $b$ so it will not be a factor of $a+b$. And no factor of $b$ will be a factof so it will not be a factor of $a+b$. And the only prime factors of $3ab$ are either prime factors of $a$, or prime factors of $b$ or $3$. No prime factor of $a$ or of $b$ are factors of $a+b$ so the only possible prime factors of $3ab$ and $a+b$ are (possibly) $3$. Or there are no common prime factors of $3ab$ and $a+b$.

If there are no common prime factors then $\gcd(a+b,3ab) = 1$.

If the only common prime factor is $3$ then $3$ is a factor of $a+b$ but not of either $a$ nor of $b$. So $\gcd(a+b, 3ab) =3$.

fleablood
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