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ganeshie8
 one year ago
There seems to be a simple+elegant way and a somewhat ugly way to solve this
Find all pairs of positive integers \((m,n)\) such that : \[\large m+n=m^2−3mn+2n^2\]
ganeshie8
 one year ago
There seems to be a simple+elegant way and a somewhat ugly way to solve this Find all pairs of positive integers \((m,n)\) such that : \[\large m+n=m^2−3mn+2n^2\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we need to put value ad try :O

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2the solution pairs (m,n) are  (0,0) (15,10) (15,12) (20,12) (14,10) (1,0) (1,2) (6,2)

sepeario
 one year ago
Best ResponseYou've already chosen the best response.1does it have something to do with factoring? because you can factor the right side.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am thinking of boundedness.. anyone got something related to that?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4i want to rearrange \(\Large m+n=m^2−3mn+2n^2\\ \Large m^22mn+n^2=m+n+mnn^2\\ \Large (mn)^2=m+n+mnn^2\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4\( \Large m^22mn+n^2=m+n+mnn^2\\ \) since we wanna positive terms then \(\Large 2nm=n^2\\\Large 2m=n \) first equation now \(\Large m^2+n^2=m+n+mn\) apply n=2m \(\Large m^2+4m^2=m+2m+2m^2 \\ \Large 3m^23m=0 \\\Large 3m(m1) =0 \\ \Large m=0 ~ or ~m=1 \) set of solutions \((m,n)\) (0,0) (1,2)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4now apply m=n/2 n^2/4 +n^2=n/2+n+n^2/2 n^2+4 n^2=2n+4 n + 2 n^2 3n^26n =0 3n(n2)=0 n=0 n=2 same solution lol ok this is all i got xD

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4ok this is the second ugly way after trial or programming .

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4ok i made bloody mistake .

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4oh wait no i didnt xD

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4still trying to match @imqwerty

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2wait we have to give only positive integer pairs so the pairs that i got are  (1,0) (1,2) (6,2)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4(1,0) is trivial but why it seems i cant get it ? also 6,2 .

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2i like such questions ( ͡° ͜ʖ ͡°) :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4me too btw this is not my q, this is part of an awesome tutorial given by @mukushla couple of years ago

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4he was here while ago, anyway whats ur two methods :O

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Here is the elegant method http://openstudy.com/updates/50028852e4b0848ddd669ca2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Amazing solution! It's a matter of shame that I couldn't think of it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2But that solution doesn't cover (6,2), does it?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4what is the ugly way ??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh good catch... @mukushla we found a mistake! plugging n=2 in the original equation gives two different values for m

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4its not a mistake as such... just another case was overlooked in the very end..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0very right :) solutions are\[(m,n)=(6, 2), (1, 2)\]

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.4ok help me with my question http://openstudy.com/updates/557981b1e4b0e4e582a9d83c
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