If $\gcd(a, b) = 1$ then prove that $\gcd(a+b, a^2-ab+b^2) = 1$ or $3$?

Asked Apr 18 '14 at 20:27

Active Jan 15 '17 at 19:53

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If $\gcd(a, b) = 1$ then prove that $\gcd(a+b, a^2-ab+b^2) = 1$ or $3$?

I am shamelessly asking how to solve the problem? I have no idea how to start and solve. Please help.

edited Jan 14 '17 at 18:05

Martin Sleziak

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asked Apr 18 '14 at 20:27

Anirban Nag 'tintinmj'

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http://math.stackexchange.com/questions/257392/if-gcda-b-1-then-gcdab-a2-abb2-1-or-3 or – lab bhattacharjee Apr 19 '14 at 04:28
And also If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. and other posts linked there. – Martin Sleziak Jan 14 '17 at 18:17

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