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Yes, I know that this questions has two answers, but I can't see why $k$ can't be 27, or any other $3^n$ with $n \neq 2$.

Community
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David
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    Because, as the accepted answer goes on to say, then $(a,b)$ would be greater than $1$, contradicting our assumption. Specifically, if $3^n\mid k$ for some $n\gt 1$, then $3^n\mid 3ab$, and therefore $3^{n-1}\mid ab$, leading only to $3\mid a$ or $3\mid b$. But $k\mid a+b$, and if $3\mid a$ or $3\mid b$ and also $3\mid a+b$, then we have $3\mid a$ and $3\mid b$ simultaneously. – abiessu Apr 06 '15 at 21:46

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Here is a simpler proof that makes it clear why that cannot occur.

${\rm mod}\ k\!:\ b\!+\!a\equiv 0\,\Rightarrow\, b\equiv -a\,\Rightarrow\, 0\equiv a^2\!+b^2\!-ab\equiv 3a^2,\, $ so $\ \color{#0a0}{k\mid 3a^2}$

But $\ k\mid a\!+\!b\,\Rightarrow\, \color{#c00}{(k,a)} = (k,a,a\!+\!b) = (k,a,b) \color{#c00}{= 1},\,$ by $\ (a,b)= 1$

Therefore, by Euclid's Lemma, $\ \color{#c00}{(k,a)=1},\ \color{#0a0}{k\mid 3a^2}\,\Rightarrow\, k\mid 3\quad $ QED

Bill Dubuque
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Following the accepted answer if $3^n|\gcd(a+b,3ab)$ with $n\gt 2$ then since $9\mid 3^n$ therefore $9|\gcd(a+b,3ab)$... now proceed according to the answer...

abiessu
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Fermat
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  • I think that your argument tell me the same: $9 \nmid a+b$ but how this imply that e.g., $27 \nmid ab$? – David Apr 06 '15 at 22:56