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Let $a$ and b be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)=1$ or $5$

Proof: $s\mid (a+b)$ and $s\mid (a^2-3ab+b^2)$ implies $s\mid(a+b)^2=a^2+b^2+2ab$ and $s\mid (a^2-3ab+b^2)$ implies $s\mid((a+b)^2-(a^2-3ab+b^2))=5ab$.

From a previous question I posted I figure that the I am on the wrong track. Can someone help me with the next step?

Martin Sleziak
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Username Unknown
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2 Answers2

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Note that $$a^2-3ab+b^2=(a+b)^2-5ab$$ So if $p$ is a prime, and such that $p|(a+b)$ and $p|(a+b)^2-5ab$ then $p|5ab$. Now $p\not | ab$ since if $p|a$, say then from $p|a+b$ follows $p|b$ contradiction the fact the $a$ and $b$ are relatively prime therefore $p|5$. The same argument shows that $gcd =5$.

Rene Schipperus
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Hint $\,\ {\rm mod}\ a\!+\!b\!:\,\ a\equiv -b\,\Rightarrow\, \color{#0a0}{a^2\!-\!3ab\!+\!b^2\equiv 5b^2}\,$ so, by the Euclidean Algorithm $\rm\color{#c00}{(EA)}$

$\quad (a\!+\!b,\color{#0a0}{a^2\!-\!3ab\!+\!b^2})\overset{\rm\color{#c00}{EA}} = (a\!+\!b,\color{#0a0}{5b^2}) = (a\!+\!b,5)\ \ {\rm by}\ \ \Bigg\{ \begin{eqnarray} &&(a\!+\!b,b)\overset{\rm\color{#c00}{EA}}= (a,b)=1\\ \Rightarrow &&(a\!+\!b,b^2) = 1\end{eqnarray}\ \,$ by Euclid.

Here $\rm\color{#c00}{EA}$ means $\ (m,n)\, =\, (m,n')\ $ if $\ n'\!\equiv n\pmod m\,\ $ [= descent step of Euclidean Algorithm].

Bill Dubuque
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