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Zaara
 2 years 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
Zaara
 2 years 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

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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4i 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

Zaara
 2 years ago
Best ResponseYou've already chosen the best response.0then how to show gcd equals 1 or 3??????

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4.... still looking at it :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4wondering if a euclidean algorithm would pan out a^2+ab+b^2 = (ab)(q) + r ...

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1not 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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0STOP RIGHT THERE. http://openstudy.com/study#/updates/512f5747e4b098bb5fbd07fa

Zaara
 2 years ago
Best ResponseYou've already chosen the best response.0better if the whole thing is explained....

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4a^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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4\[cd~\iff~d=cn~;~n \in Z\] \[p = kq+r~\to \frac{pr}{k}=q~,~q\in Z\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4the euclidean algorithm has alot of recurrsive:\[a=bq+r\] if \[a=bq+r\\b = r(b)+0\]then gcd(a,b)=r

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4hmmm, 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 :)

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.4pffft, misread my own thoughts lol if b=rb then a = rbq+r a = r(bq+1) for q in Z
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