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anonymous

  • 5 years ago

Hi everybody! Find all positive integral solutions (x,y) of the polynomial equation: 4x^3+4x^2y-15xy^2-18y^3-12x^2+6xy+36y^2+5x-10y=0

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  1. anonymous
    • 5 years ago
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    any thoughts/ideas?

  2. angela210793
    • 5 years ago
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    couldn't u find a longer eq? :P

  3. anonymous
    • 5 years ago
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    its a monstrous beast...

  4. angela210793
    • 5 years ago
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    :P

  5. angela210793
    • 5 years ago
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    I think it is smth like (ax-by)^3 bu i can't tell wht r a and b

  6. angela210793
    • 5 years ago
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    but*

  7. anonymous
    • 5 years ago
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    good try angela..but u never know that will factorize completely to 3 linear factors!... Infact this BEAST can be factored to any one of the following: 1.3 linears. 2.1 linear and 1 conic

  8. anonymous
    • 5 years ago
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    (x-2*y)*(2*x-1+3*y)*(2*x-5+3*y)

  9. anonymous
    • 5 years ago
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    (x-2y)(2x+3y-5)(2x+3y-1)=0

  10. angela210793
    • 5 years ago
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    i'm not planning to deal with this..u do it...:P

  11. anonymous
    • 5 years ago
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    i got it first, please medal

  12. anonymous
    • 5 years ago
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    u rock cantorset! but we cant use comps!

  13. anonymous
    • 5 years ago
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    how do you know i use computer

  14. anonymous
    • 5 years ago
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    :)

  15. anonymous
    • 5 years ago
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    @contorset: lol, its obvious. :P

  16. angela210793
    • 5 years ago
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    hey how did u do tht?

  17. anonymous
    • 5 years ago
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    well witout at computer its LUCK..

  18. anonymous
    • 5 years ago
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    I didn't do it on my own, wolframalpha

  19. anonymous
    • 5 years ago
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    maple

  20. angela210793
    • 5 years ago
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    and how do u do it in comp?

  21. anonymous
    • 5 years ago
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    haha... if u had a good luck u wud put x=2y and inspect!!!

  22. anonymous
    • 5 years ago
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    you can do it in wolframalpha,

  23. anonymous
    • 5 years ago
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    then u get (x-2y) and get others by dividing

  24. anonymous
    • 5 years ago
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    i should not give out my secrets though

  25. anonymous
    • 5 years ago
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    k. but tat doesnt solve the problem..

  26. anonymous
    • 5 years ago
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    x=2y ? you mean substitute

  27. anonymous
    • 5 years ago
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    @saubhik: You need a VERY good luck for that!

  28. anonymous
    • 5 years ago
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    yes...

  29. anonymous
    • 5 years ago
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    but see tat doesnt..solve the problem

  30. anonymous
    • 5 years ago
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    we have to find positive integer solutions..

  31. angela210793
    • 5 years ago
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    woww

  32. anonymous
    • 5 years ago
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    but you can pick anything, x = 4y for instance

  33. angela210793
    • 5 years ago
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    and i thought u guys were smart :P:P:P

  34. anonymous
    • 5 years ago
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    i dont see any logic in this problem

  35. anonymous
    • 5 years ago
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    ya cantorset that's truly possible but as i said we need luck and brain-breaking insights...

  36. anonymous
    • 5 years ago
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    see..now how to find the solution?

  37. anonymous
    • 5 years ago
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    for the first one(x-2y), x = 2I y=I

  38. anonymous
    • 5 years ago
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    u can have infinite pairs (2n,n) for a positive integer n since x-2y is a factor...what for the other two?

  39. anonymous
    • 5 years ago
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    yes

  40. anonymous
    • 5 years ago
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    what for 2x+3y-5=0

  41. anonymous
    • 5 years ago
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    its obvious (1,1) but are there others..THIS IS DIOPHANTINE!!

  42. anonymous
    • 5 years ago
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    OK lets see, x + 3y/2 =5/2 x=(5-3y)/2 5-3y should be even,

  43. anonymous
    • 5 years ago
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    so y is odd.

  44. anonymous
    • 5 years ago
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    what about 2x+3y=1? yay...theres no positive integers here in this equation..so this is infertile:P

  45. anonymous
    • 5 years ago
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    Why do you think so?

  46. anonymous
    • 5 years ago
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    how can u get postive integers (x,y) from 2x+3y<5

  47. anonymous
    • 5 years ago
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    there are none...:P

  48. anonymous
    • 5 years ago
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    infertile..mother has no babies:P

  49. anonymous
    • 5 years ago
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    that can be solved with euclids algorithm

  50. anonymous
    • 5 years ago
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    Oh sorry, i dint read that 'positive' part! :P

  51. anonymous
    • 5 years ago
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    this mother 2x+3y=5 has only one baby i.e. (1,1)

  52. anonymous
    • 5 years ago
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    So our answer: (1,1) and (2n,n)

  53. angela210793
    • 5 years ago
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    OMG!!! This is giving me headache only by reading ur explanations imagine if i had to solve this....:@@@@@@@@

  54. anonymous
    • 5 years ago
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    So its done, (n,2n) U (1,1)

  55. anonymous
    • 5 years ago
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    Openstudy rocks!! thnx evrybody!

  56. anonymous
    • 5 years ago
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    Oops (2n,n). yes right

  57. angela210793
    • 5 years ago
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    Going to deal with literature bye and have a good time :)

  58. anonymous
    • 5 years ago
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    , amog, you wrote x= 1, y = 21?

  59. anonymous
    • 5 years ago
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    @angela: buh-bye. @cantorset: It was an 'I' (eye)

  60. anonymous
    • 5 years ago
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    for the first one(x-2y), x = 2I y=I

  61. anonymous
    • 5 years ago
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    Yes i corrected it, didn't I?

  62. anonymous
    • 5 years ago
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    hi cantorset...do u mind givin me an idea how to have a factor to the eqn..

  63. anonymous
    • 5 years ago
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    suppose i have the baddest luck and go on inspecting from x=3y;x=4y....

  64. anonymous
    • 5 years ago
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    wait, ... start over . why would i think to use x = 3y or anything like that

  65. anonymous
    • 5 years ago
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    also its 3 linear, not a conic

  66. anonymous
    • 5 years ago
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    or worser x=3y+7 etc..

  67. anonymous
    • 5 years ago
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    yeah

  68. anonymous
    • 5 years ago
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    you mean because the solution is ( ax + by ) ( dx + cy + ...

  69. anonymous
    • 5 years ago
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    See cantormath..the given equation Infact this BEAST can be factored to any one of the following: 1.3 linears. 2.1 linear and 1 conic

  70. anonymous
    • 5 years ago
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    now clearly theres no constant term in this equation...so a linear factor (which must be present) must pass through the origin...

  71. anonymous
    • 5 years ago
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    and so we start checking by x=y;x=2y;x=3y...

  72. anonymous
    • 5 years ago
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    if the question does not mean to torment us we can get the factor in a few trials...

  73. anonymous
    • 5 years ago
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    k..thnx evrybody for helping..:)

  74. anonymous
    • 5 years ago
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    wait, youre saying in general, a cubic in x and y can be factored into either 3 linears, or 1 linear and 2 conics? , or 2 linears, and a conic?

  75. anonymous
    • 5 years ago
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    what do you mean by constant term in the equation

  76. anonymous
    • 5 years ago
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    term with no variables attched..

  77. anonymous
    • 5 years ago
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    oh right

  78. anonymous
    • 5 years ago
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    the final solution is 3 linears, right?

  79. anonymous
    • 5 years ago
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    yups...family of 3 straight line

  80. anonymous
    • 5 years ago
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    cool, so ok , theres no constant term, and clearly, (0,0) works

  81. anonymous
    • 5 years ago
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    yups..try looking for x=y,2y,3y,...here u need luck

  82. anonymous
    • 5 years ago
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    so that tells us that x = ay or y = ax must be one of the solutions?

  83. anonymous
    • 5 years ago
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    yes..true

  84. anonymous
    • 5 years ago
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    but only for a=2

  85. anonymous
    • 5 years ago
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    i.e.(2n,n)

  86. anonymous
    • 5 years ago
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    right, but thats only if you do the long division< right?

  87. anonymous
    • 5 years ago
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    if you know to assume, 2x = y ?

  88. anonymous
    • 5 years ago
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    or wait, you know that (0,0) is a solution and (1,1) is a solution, right>

  89. anonymous
    • 5 years ago
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    yes...not assume but see if its correct by replacing 2x by y OR replacing y by 2x

  90. anonymous
    • 5 years ago
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    (1,1) is known after we get other factors using x-2y

  91. anonymous
    • 5 years ago
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    hmmm, yes but i mean is there something that would make us want to to do x = 2y

  92. anonymous
    • 5 years ago
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    oh ,

  93. anonymous
    • 5 years ago
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    tat is LUCK!

  94. anonymous
    • 5 years ago
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    ( 1,-1) is another solution then

  95. anonymous
    • 5 years ago
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    OR if u knew the problem beforehand OR if u have wolframalpha open on ur browser. No...-1 is not a POSITIVE integer :P

  96. anonymous
    • 5 years ago
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    4x^2 + 9y^2 + 12xy -2x -15y +5 =0 How do you solve this btw?

  97. anonymous
    • 5 years ago
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    from discriminant criteria for various conics see if its parabola/ellipse/hyperbola..

  98. anonymous
    • 5 years ago
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    k. for positive integer solutions...ALL 2-VARIABLE EQUATION DO NOT HAVE positive integer solutions.....such questions are given after testing them..

  99. anonymous
    • 5 years ago
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    OK!

  100. anonymous
    • 5 years ago
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    amogh u indian?

  101. anonymous
    • 5 years ago
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    So here its a parabola

  102. anonymous
    • 5 years ago
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    tats what wolframalpha says...

  103. anonymous
    • 5 years ago
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    I didn't use wolframalpha for it? Afterall we in India are not allowed to use any calc! :P

  104. anonymous
    • 5 years ago
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    And I like it!

  105. anonymous
    • 5 years ago
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    in which class r u?

  106. anonymous
    • 5 years ago
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    Just went into XII

  107. anonymous
    • 5 years ago
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    wer r u frm?

  108. anonymous
    • 5 years ago
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    kolkata passed 12

  109. anonymous
    • 5 years ago
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    what are the possible shapes of a third degree x,y equation, a conic is second degree,

  110. anonymous
    • 5 years ago
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    Cool, did you give the JEE?

  111. anonymous
    • 5 years ago
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    may be: 1.family of straight lines.. 2. a strt line and a conic

  112. anonymous
    • 5 years ago
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    gave it! but didnt get it:(

  113. anonymous
    • 5 years ago
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    hope u get:)

  114. anonymous
    • 5 years ago
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    Thanks :) So what plans?

  115. anonymous
    • 5 years ago
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    very cool saubhik, didnt know that

  116. anonymous
    • 5 years ago
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    dont know ....awaiting for AIEEE and WBJEE and ISI results...

  117. anonymous
    • 5 years ago
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    whats AIEEE?

  118. anonymous
    • 5 years ago
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    All India Engineering Entrance Examination (in india)

  119. anonymous
    • 5 years ago
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    thnx..cantorset...

  120. anonymous
    • 5 years ago
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    All the best buddy!

  121. anonymous
    • 5 years ago
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    saubhik, i got a problem for ya

  122. anonymous
    • 5 years ago
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    mayb ei should post new one

  123. anonymous
    • 5 years ago
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    yups...

  124. anonymous
    • 5 years ago
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    post new one..

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