Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

i wonder a-b = c a^2+ab+b^2 = d a-b+ab = c +ab a^2+ab+ab+b^2 = d +ab a-b+ab = c +ab (a+b)^2 = d +ab
then how to show gcd equals 1 or 3??????

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

.... still looking at it :)
wondering if a euclidean algorithm would pan out a^2+ab+b^2 = (a-b)(q) + r ...
i am going to make a guess that it has something to do with \[a^3-b^2=(a-b)(a^2+ab+b^2)\] just a guess though
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
@ParthKohli thats it!!!
better if the whole thing is explained....
a^2+ab+b^2 = (a-b)(q) + r a^2+ab+b^2 - r= (a-b)(q) (a^2+ab+b^2 - r)(a-b) = q i was wondering if there was a method we could employ there; give that r=1 or 3, and gcd(a,b) = 1
*bookmark*
:-\ \[\dfrac{(a^2 + ab + b^2 - r)}{a - b} = q\]
Can anyone explain?
\[c|d~\iff~d=cn~;~n \in Z\] \[p = kq+r~\to \frac{p-r}{k}=q~,~q\in Z\]
the euclidean algorithm has alot of recurrsive:\[a=bq+r\] if \[a=bq+r\\b = r(b)+0\]then gcd(a,b)=r
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: \[\frac{a^2+ab+b^2}{(a-b)^2+1}=1~or~3\]should do it :)
pffft, misread my own thoughts lol if b=rb then a = rbq+r a = r(bq+1) for q in Z

Not the answer you are looking for?

Search for more explanations.

Ask your own question