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if a and b are two integers with gcd(a,b)=1, then show that gcd(ab, a^2+ab+b^2)=1 or 3
 one year ago
 one year ago
if a and b are two integers with gcd(a,b)=1, then show that gcd(ab, a^2+ab+b^2)=1 or 3
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.4
i wonder ab = c a^2+ab+b^2 = d ab+ab = c +ab a^2+ab+ab+b^2 = d +ab ab+ab = c +ab (a+b)^2 = d +ab
 one year ago

ZaaraBest ResponseYou've already chosen the best response.0
then how to show gcd equals 1 or 3??????
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
.... still looking at it :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
wondering if a euclidean algorithm would pan out a^2+ab+b^2 = (ab)(q) + r ...
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i am going to make a guess that it has something to do with \[a^3b^2=(ab)(a^2+ab+b^2)\] just a guess though
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
not a solution by any means, i just thought it might be the hook in to the problem i have no idea how to solve it
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
STOP RIGHT THERE. http://openstudy.com/study#/updates/512f5747e4b098bb5fbd07fa
 one year ago

ZaaraBest ResponseYou've already chosen the best response.0
better if the whole thing is explained....
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
a^2+ab+b^2 = (ab)(q) + r a^2+ab+b^2  r= (ab)(q) (a^2+ab+b^2  r)(ab) = q i was wondering if there was a method we could employ there; give that r=1 or 3, and gcd(a,b) = 1
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
:\ \[\dfrac{(a^2 + ab + b^2  r)}{a  b} = q\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Can anyone explain?
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
\[cd~\iff~d=cn~;~n \in Z\] \[p = kq+r~\to \frac{pr}{k}=q~,~q\in Z\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
the euclidean algorithm has alot of recurrsive:\[a=bq+r\] if \[a=bq+r\\b = r(b)+0\]then gcd(a,b)=r
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
hmmm, so if we can show that: a = r(b^2+1) or a/(b^2+1) = r it looks good to me :) is there a flaw in my thought perhaps? if not then: \[\frac{a^2+ab+b^2}{(ab)^2+1}=1~or~3\]should do it :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.4
pffft, misread my own thoughts lol if b=rb then a = rbq+r a = r(bq+1) for q in Z
 one year ago
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