Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Zaara
Group Title
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
Zaara Group Title
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

This Question is Open

Zaara Group TitleBest ResponseYou've already chosen the best response.0
@experimentX
 one year ago

amistre64 Group TitleBest 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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.4
.... still looking at it :)
 one year ago

amistre64 Group TitleBest 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

satellite73 Group TitleBest 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

satellite73 Group TitleBest 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

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

Zaara Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli thats it!!!
 one year ago

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

amistre64 Group TitleBest 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

Callisto Group TitleBest ResponseYou've already chosen the best response.0
*bookmark*
 one year ago

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

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

amistre64 Group TitleBest 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

amistre64 Group TitleBest 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

amistre64 Group TitleBest 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

amistre64 Group TitleBest 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
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.