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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 : \[\large m+n=m^2−3mn+2n^2\]

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  1. anonymous
    • one year ago
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    we need to put value ad try :O

  2. imqwerty
    • one year ago
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    the solution pairs (m,n) are - (0,0) (-15,-10) (-15,-12) (-20,-12) (-14,-10) (1,0) (1,2) (6,2)

  3. sepeario
    • one year ago
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    does it have something to do with factoring? because you can factor the right side.

  4. anonymous
    • one year ago
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    i am thinking of boundedness.. anyone got something related to that?

  5. ikram002p
    • one year ago
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    i want to rearrange \(\Large m+n=m^2−3mn+2n^2\\ \Large m^2-2mn+n^2=m+n+mn-n^2\\ \Large (m-n)^2=m+n+mn-n^2\)

  6. ikram002p
    • one year ago
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    \( \Large m^2-2mn+n^2=m+n+mn-n^2\\ \) 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)

  7. ikram002p
    • one year ago
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    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

  8. ikram002p
    • one year ago
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    ok this is the second ugly way after trial or programming -.-

  9. ikram002p
    • one year ago
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    ok i made bloody mistake -.-

  10. ikram002p
    • one year ago
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    oh wait no i didnt xD

  11. ikram002p
    • one year ago
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    still trying to match @imqwerty

  12. imqwerty
    • one year ago
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    wait we have to give only positive integer pairs so the pairs that i got are - (1,0) (1,2) (6,2)

  13. ikram002p
    • one year ago
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    (1,0) is trivial but why it seems i cant get it ? also 6,2 -.-

  14. ganeshie8
    • one year ago
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    Nice!

  15. imqwerty
    • one year ago
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    i like such questions ( ͡° ͜ʖ ͡°) :)

  16. ganeshie8
    • one year ago
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    me too btw this is not my q, this is part of an awesome tutorial given by @mukushla couple of years ago

  17. ikram002p
    • one year ago
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    he was here while ago, anyway whats ur two methods :O

  18. ganeshie8
    • one year ago
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    Here is the elegant method http://openstudy.com/updates/50028852e4b0848ddd669ca2

  19. ParthKohli
    • one year ago
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    Amazing solution! It's a matter of shame that I couldn't think of it.

  20. ParthKohli
    • one year ago
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    But that solution doesn't cover (6,2), does it?

  21. ikram002p
    • one year ago
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    what is the ugly way ??

  22. ganeshie8
    • one year ago
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    Ahh good catch... @mukushla we found a mistake! plugging n=2 in the original equation gives two different values for m

  23. ParthKohli
    • one year ago
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    :)

  24. anonymous
    • one year ago
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    Let me see :))

  25. ganeshie8
    • one year ago
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    its not a mistake as such... just another case was overlooked in the very end..

  26. anonymous
    • one year ago
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    very right :-) solutions are\[(m,n)=(6, 2), (1, 2)\]

  27. ikram002p
    • one year ago
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    ok help me with my question http://openstudy.com/updates/557981b1e4b0e4e582a9d83c

  28. anonymous
    • one year ago
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    ...

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