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we need to put value ad try :O
the solution pairs (m,n) are - (0,0) (-15,-10) (-15,-12) (-20,-12) (-14,-10) (1,0) (1,2) (6,2)
does it have something to do with factoring? because you can factor the right side.
i am thinking of boundedness.. anyone got something related to that?
i want to rearrange \(\Large m+n=m^2−3mn+2n^2\\ \Large m^2-2mn+n^2=m+n+mn-n^2\\ \Large (m-n)^2=m+n+mn-n^2\)
\( \Large m^2-2mn+n^2=m+n+mn-n^2\\ \) 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)
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
ok this is the second ugly way after trial or programming -.-
ok i made bloody mistake -.-
oh wait no i didnt xD
still trying to match @imqwerty
wait we have to give only positive integer pairs so the pairs that i got are - (1,0) (1,2) (6,2)
(1,0) is trivial but why it seems i cant get it ? also 6,2 -.-
i like such questions ( ͡° ͜ʖ ͡°) :)
me too btw this is not my q, this is part of an awesome tutorial given by @mukushla couple of years ago
he was here while ago, anyway whats ur two methods :O
Amazing solution! It's a matter of shame that I couldn't think of it.
But that solution doesn't cover (6,2), does it?
what is the ugly way ??
Ahh good catch... @mukushla we found a mistake! plugging n=2 in the original equation gives two different values for m
Let me see :))
its not a mistake as such... just another case was overlooked in the very end..
very right :-) solutions are\[(m,n)=(6, 2), (1, 2)\]