Question

Please help:)
if a and b are two integers with g.c.d(a,b)=1, then show that g.c.d(a-b, a^2+ab+b^2)=1 or 3

Answer 1

Let d=g.c.d(a-b,a^2+ab,b^2). We want to show that d=1 or 3.
Since d is a divisor of a-b it is also a divisor of (a-b)^2 and -(a-b)^2.
Since d is a divisor of -(a-b)^2 and a^2+ab+b^2
it is a divisor of their sum.
Sum: -(a-b)^2+a^2+ab+b^2=3ab
Am I right upto this part?

Answer 2

Hello ajprincess

quite right till there..

Answer 3

Hello. Thanx @mukushla. Hw will I prove d=1 or 3 after this?

Answer 4

=> d|3ab
=> d|3a and d|b or d|a and d|3b
Thus d=1 or 3
I am nt sure if I am right

Answer 5

guys, lend a hand :) what do u think?

Answer 6

You know there is a saying, if you don't know much about something, better keep mum and let others wonder whether you know it or not, rather than opening your mouth and letting others know!
I'd like to follow that principle! :)

Answer 7

i think you need to show that d is divisor of ab also.

Answer 8

then since d is divior of ab and 3ab, d=1 or 3

Answer 9

do u know how can u prove that di is divisor of ab ?

Answer 10

hmm no.

Answer 11

d is divisor of (a-b)^2 and a^2+ab+b^2
take difference this time...

Answer 12

i was doing that in mind...so just a sec...

Answer 13

you'll again get 3ab :P

Answer 14

ya it will be -3ab this time

Answer 15

wait for a second friends ...

Answer 16

suppose d|a, because d|a-b then d|b
d|a, d|b, d^2|ab
d is the greatest, and d^2|ab then d=1

Answer 17

u may do it like this

\(d|a-b\) and \(d|3ab\) so \(d|3ab-3b(a-b)=3b^2\) using same procedure

\(d|a-b\) and \(d|3ab\) so \(d|3ab+3a(a-b)=3a^2\) finally

\(d|\gcd(3a^2,3b^2)=3\gcd(a^2,b^2)=3\)

Answer 18

Sorry. I dnt get the part hw u got d|3ab-3b(a-b)

Answer 19

Given : (a,b) = 1

To prove : \(\large{(a-b),(a^2+b^2+ab)= 1 \space \textbf{or} \space 3}\)

proof : (a,b) = 1 , let : a = bq + r , \(\large{0 < r < q }\) and (b,r) = 1 are satisfied

a - b = bq + r - b = b(q-1) + r

let \(\large{(a-b),(a^2+b^2+ab) \ne 1}\)
=> \(\large{a-b|a^2 + b^2 + ab}\) --------------1)
Also , \(\large{a-b | (a-b)^2}\) => \(\large{a-b|a^2+b^2-2ab}\) ----2)
From 1) and 2) : \(\large{a-b|a^2 + b^2 + ab - a^2 -b^2 +2ab \implies a-b |3ab}\)

Now , \(\large{a-b \space \textbf{not divide} \space a}\) and \(\large{a-b \space \textbf{not divide} \space b}\)

therefore \(\large{a-b | 3}\)
Hence \(\large{ (a-b, a^2+b^2+ab) = 3}\) iff it is not equal to 1

now, if \(\large{a-b \space \textbf{not divide} \space 3}\) then :
a-b not divide 3ab

therefore a-b not divides (a-b)^2 + 3ab

implies, a-b not divides (a^2 +b^2 + ab)

that is (a-b, a^2+b^2+ab) = 1
proved

Answer 20

@ajprincess when d is divisor of both a and b then d is also a divisor of every linear combination of a and b like am+bn

Answer 21

Is my method correct ?

Answer 22

Sorry @mathslover I havnt any idea if ur method is right. i got it @mukushla. so first we need to show d is a divisor of a and b right?

Answer 23

no, i think, what he meant was,
if d is divisor of A and B, then
d is also divisor of A and -3b(B)
here A =3ab, B=a-b

Answer 24

np , even I came to the above solution after thinking a lot... May be you didn't get it because I posted the solution in a single post and didn't go through step by step and making you understand that by discussing, I was late and so I thought I should post it in a single post now. Sorry but nice work @mukushla

Answer 25

Thanx a lot @mathslover, @mukushla nd @hartnn