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anonymous

  • one year ago

How many solutions which integers such that x^2 + y^2 + z^2 = 2016

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  1. ganeshie8
    • one year ago
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    I think it is sufficient to consider solutions in positive integers first, then we can build the complete solutions by permutations.

  2. ganeshie8
    • one year ago
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    Working mod 4 it is easy to see that each of x, y, z must be of form \(4k\). Otherwise left hand side is not divisible by 16. So the equation reduces to \[a^2+b^2+c^2 = 126\] where \(x = 4a\), \(y=4b\), \(z=4c\)

  3. anonymous
    • one year ago
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    apparently gauss did it over 200 years ago but hmm

  4. ganeshie8
    • one year ago
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    related to quadratic reciprocity law in any ways ? maybe taking mod 7 or 3 might narrow it further idk... my attempt above looks very hacky..

  5. anonymous
    • one year ago
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    actually, he did it for square free numbers, so I guess we're screwed

  6. ganeshie8
    • one year ago
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    looks interesting http://gyazo.com/301173b00539b4577e227beafd0b4281

  7. dan815
    • one year ago
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    |dw:1436766690043:dw|

  8. dan815
    • one year ago
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    |dw:1436766915671:dw|

  9. dan815
    • one year ago
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    |dw:1436767224913:dw|

  10. dan815
    • one year ago
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    lets start by considering the total number of ordered pairs of inteegers in this arc

  11. dan815
    • one year ago
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    then we can see which one of them will produce an int for the z coordinate too

  12. ganeshie8
    • one year ago
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    There is a nice formula for number of representations of \(n\) as sum of two squares though

  13. ganeshie8
    • one year ago
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    \[x^2+y^2 = 2016\] has no solutions over integers

  14. dan815
    • one year ago
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    |dw:1436767447283:dw|

  15. dan815
    • one year ago
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    this will give us a max bound on the number of integer solutions atleast

  16. ganeshie8
    • one year ago
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    we can discard 2016 and work on smaller number 126 because both equations must have same number of solutions

  17. dan815
    • one year ago
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    how come we can do that?

  18. ganeshie8
    • one year ago
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    \(x^2+y^2+z^2=2016 = 16*126 \) consider this equation in mod 4 suppose \(x\equiv y\equiv z\equiv 0 \pmod{4}\) then \(x^2+y^2+z^2 = 16(a^2+b^2+c^2) = 2016 \implies a^2+b^2+c^2 = 126\)

  19. dan815
    • one year ago
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    but we dont have to be able to factor 16 from all the squares

  20. ganeshie8
    • one year ago
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    left hand side wont be divisible by \(16\) for all other cases, so we don't need to wry about them

  21. ganeshie8
    • one year ago
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    for example, \(x\equiv y\equiv z\equiv 1\pmod{4}\) makes left hand side \(\equiv 3 \pmod{16}\) but the right hand side is \(\equiv 0 \pmod{16}\) similarly you can show that all the remaining cases don't work

  22. dan815
    • one year ago
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    what if we are dealing with cases where x y and z are all different mod and end up adding to 0 mod 16

  23. ganeshie8
    • one year ago
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    thats exactly what im saying, thats not possible. maybe consider another example : \(x\equiv 2, y\equiv 1,z\equiv 0 \pmod{4}\) then \[x^2\equiv 0,~y^2\equiv 1, ~z^2\equiv 0\] clearly left hand side is \(\equiv 1 \pmod{4}\) but the right hand side is \(\equiv 0\pmod{4}\)

  24. dan815
    • one year ago
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    okk i see what u are saying

  25. dan815
    • one year ago
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    okay one more question, dont we have to consider the behavior in other mods too?

  26. ganeshie8
    • one year ago
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    Nope, any integer can be represented in any one of the form \(4k\pm a\) where \(0\le a\lt 4\). s we're done if we're done in mod 4

  27. ganeshie8
    • one year ago
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    this is same as considering even/odd cases right

  28. ganeshie8
    • one year ago
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    If we prove/disprove an integer equation for both odd and even cases, then we're done. working even/odd cases is same as mod2, we dont need to wry about other mods

  29. dan815
    • one year ago
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    ok isee

  30. dan815
    • one year ago
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    can we continue the process then

  31. ganeshie8
    • one year ago
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    Any square number is either \(0\) or \(1\) in \(\mod 4\), and clearly \(X+Y+Z \not\equiv 0\) when at least one of the variables on left hand side is \(1\). so it must be the case that \(X\equiv Y\equiv Z\equiv 0\)

  32. dan815
    • one year ago
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    and factor 3^2 and solve in mode 3?

  33. ganeshie8
    • one year ago
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    actually i was stuck with \[a^2+b^2+c^2 = 126\] mod 3 is not making it any simpler

  34. dan815
    • one year ago
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    11,2,1

  35. dan815
    • one year ago
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    the simplest one i can find

  36. ganeshie8
    • one year ago
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    Finding each solution gives us 3! = 6 solutions by permuting the ordered pairs

  37. dan815
    • one year ago
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    the choices are pretty low to pick from, maybe they expect brute force from here

  38. ganeshie8
    • one year ago
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    yeah

  39. dan815
    • one year ago
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    11 is the highest 10 is ruled out 100 + 26 x^2+y^2=26 has no inter solution 9 is ruled out too 81 + 35 x^2+y^2=35 no int solution

  40. dan815
    • one year ago
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    oh wait 25+1

  41. dan815
    • one year ago
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    10 is good too

  42. dan815
    • one year ago
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    nothing else

  43. dan815
    • one year ago
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    11,2,1 10,5,1

  44. dan815
    • one year ago
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    checking 9 81+45 45=6^2+3^2

  45. dan815
    • one year ago
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    9,6,3

  46. ganeshie8
    • one year ago
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    thats all

  47. dan815
    • one year ago
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    11,2,1 10,5,1 9,6,3 and so on

  48. dan815
    • one year ago
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    theres gotta be a better way than brute force

  49. ganeshie8
    • one year ago
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    there is a better way when the right hand side is squarefree, but 126 is not square free. so there is no known simple method

  50. dan815
    • one year ago
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    theres nothing with 8 and 7, so we are actually done here

  51. dan815
    • one year ago
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    since it will repeat

  52. dan815
    • one year ago
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    11,2,1 10,5,1 9,6,3 so 6*5*4 *3 right?

  53. ganeshie8
    • one year ago
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    3*3! positive solitions 3*3!*2^3 integer solutions ?

  54. dan815
    • one year ago
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    i was thinking well its plus or minus 11,2,1 so we have 6 choices and 3 spots 6*5*4 so 6*5*4 for each triple ordered pair

  55. dan815
    • one year ago
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    oh oops ya you are right, if i do this ill allow -11 and 11

  56. dan815
    • one year ago
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    (6*4*2) * 3 = 3*3! * 2^3 oh this works too

  57. dan815
    • one year ago
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    |dw:1436769950610:dw|

  58. ganeshie8
    • one year ago
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    yes 144 looks correct!

  59. anonymous
    • one year ago
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    Yups the answer is 144. You all are best mathematiciants :)

  60. anonymous
    • one year ago
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    Thank you very much for your answer guys

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