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Solve this Maths question, or else, you aren't good enough for Oxbridge Watch

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    Before you write bs, I know the answer

    x+y+z= 1
    x^2+y^2+z^2= 39
    x^3+y^3+z^3= 61 In all respective integers
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    13.


    gg




    (honestly, I could not care, so I stated my favourite number.)
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    Easy question, just takes a lot of time to do because of all the substitutions
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    (Original post by JAIYEKO)
    Before you write bs, I know the answer

    x+y+z= 1
    x^2+y^2+z^2= 39
    x^3+y^3+z^3= 61 In all respective integers
    So, you consider yourself to be some sort of elite mathematician?
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    (Original post by JAIYEKO)
    Yes I do, problem?
    I guess you could prove yourself by solving a problem my humble mind was capable of, then, rather than giving us a problem we cannot hope to solve.

    So prove [0,1] and (0,1) are homotopy equivalent, please.
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    (Original post by FireGarden)
    ...
    Bruh, ya dun dropped it.
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    (Original post by JAIYEKO)
    Before you write bs, I know the answer

    x+y+z= 1
    x^2+y^2+z^2= 39
    x^3+y^3+z^3= 61 In all respective integers
    Damn, I got close, but my squares add up to 41 not 39

    [5, -4, 0]
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    (Original post by JAIYEKO)
    Before you write bs, I know the answer

    x+y+z= 1
    x^2+y^2+z^2= 39
    x^3+y^3+z^3= 61 In all respective integers
    -3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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    (Original post by JAIYEKO)
    I'm arguably and possibly the only person in UK who can solve such of a difficult question
    That's cute.
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    (Original post by JAIYEKO)
    Before you write bs, I know the answer

    x+y+z= 1
    x^2+y^2+z^2= 39
    x^3+y^3+z^3= 61 In all respective integers
    Not difficult at all.

    The solutions are:
    -0.5278597049, 4.910628066, -3.857842096
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    (Original post by Aph)
    -3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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    He said integers, so yo' wrong son.

    Legendres three square theorem (correct me if I'm wrong) states that no three integers can add up to 39.
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    (Original post by JAIYEKO)
    I'm arguably and possibly the only person in UK who can solve such of a difficult question
    See I told you that you are a troll.
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    There aren't integer solutions to this problem. Are you sure you didn't mean:

    x^2+y^2+z^2=41?
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    (Original post by TheFOMaster)
    He said integers
    The solution I gave is the only one. That means that if he wants x,y and z to be integers then there is no solution.

    OP, did you type the correct numbers?
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    (Original post by Forum User)
    Damn, I got close, but my squares add up to 41 not 39

    [5, -4, 0]

    The closest person yet
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    (Original post by Aph)
    -3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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    Integers, sorry
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    (Original post by fat_hampster)
    There aren't integer solutions to this problem. Are you sure you didn't mean:

    x^2+y^2+z^2=41?
    I think i know what i wrote
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    For anyone that's interested ^
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    (Original post by Aph)
    -3.858, -0.053 and 4.911 are the approx values of the triples boring btw

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    This - or one can show that it follows that xyz=1 so the roots can't be integers.
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    (Original post by RichE)
    This - or one can show that it follows that xyz=1 so the roots can't be integers.
    This one too.
 
 
 
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