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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??????

.... still looking at it :)

wondering if a euclidean algorithm would pan out
a^2+ab+b^2 = (a-b)(q) + r ...

STOP RIGHT THERE.
http://openstudy.com/study#/updates/512f5747e4b098bb5fbd07fa

@ParthKohli thats it!!!

better if the whole thing is explained....

*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

pffft, misread my own thoughts lol
if b=rb then a = rbq+r
a = r(bq+1) for q in Z