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 : $$\large m+n=m^2−3mn+2n^2$$

Question

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 :  $$\large m+n=m^2−3mn+2n^2$$

anonymous · Answer

we need to put value ad try :O

imqwerty · Answer

the solution pairs (m,n) are -
(0,0)
(-15,-10)
(-15,-12)
(-20,-12)
(-14,-10)
(1,0)
(1,2)
(6,2)

sepeario · Answer

does it have something to do with factoring? because you can factor the right side.

anonymous · Answer

i am thinking of boundedness.. anyone got something related to that?

ikram002p · Answer

i  want to rearrange

$\Large m+n=m^2−3mn+2n^2\ \Large  m^2-2mn+n^2=m+n+mn-n^2\
\Large (m-n)^2=m+n+mn-n^2$

ikram002p · Answer

$  \Large  m^2-2mn+n^2=m+n+mn-n^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^2-3m=0 \\Large 3m(m-1) =0 \ \Large  m=0 ~ or  ~m=1 $

set of solutions  $(m,n)$
(0,0)
(1,2)

ikram002p · Answer

now 
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^2-6n =0
3n(n-2)=0
n=0
n=2

same solution lol ok this is all i got xD

ikram002p · Answer

ok this is the second ugly way after trial or programming -.-

ikram002p · Answer

ok i made bloody mistake -.-

ikram002p · Answer

oh wait no i didnt xD

ikram002p · Answer

still trying to match @imqwerty

imqwerty · Answer

wait we have to give only positive integer pairs so the pairs that i got are - 
(1,0)
 (1,2) 
(6,2)

ikram002p · Answer

(1,0) is trivial but why it seems i cant get it ? also 6,2 -.-

ganeshie8 · Answer

Nice!

imqwerty · Answer

i like such questions ( ͡° ͜ʖ ͡°)  :)

ganeshie8 · Answer

me too 
btw this is not my q, this is part of an awesome tutorial given by @mukushla  couple of years ago

ikram002p · Answer

he was here while ago, anyway whats ur two methods :O

ganeshie8 · Answer

Here is the elegant method http://openstudy.com/updates/50028852e4b0848ddd669ca2

parthkohli · Answer

Amazing solution! It's a matter of shame that I couldn't think of it.

parthkohli · Answer

But that solution doesn't cover (6,2), does it?

ikram002p · Answer

what is the ugly way ??

ganeshie8 · Answer

Ahh good catch...  @mukushla  we found a mistake!
plugging n=2 in the original equation gives two different values for m

parthkohli · Answer

:)

anonymous · Answer

Let me see :))

ganeshie8 · Answer

its not a mistake as such... just another case was overlooked in the very end..

anonymous · Answer

very right :-) solutions are$$(m,n)=(6, 2), (1, 2)$$

ikram002p · Answer

ok help me with my question http://openstudy.com/updates/557981b1e4b0e4e582a9d83c

anonymous · Answer

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