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Math Help - Simultaneous Equations (BMO Past Question)

  1. #1
    Junior Member SuperCalculus's Avatar
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    Simultaneous Equations (BMO Past Question)

    So yeah, I'm looking at British Mathematical Olympiad questions, and this one has me utterly stumped. Any ideas?

    Find all integers x, y and z such that:

    x^2 + y^2 + z^2 = 2(yz + 1)
    And
    x + y + z = 4018.
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  2. #2
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    Quote Originally Posted by SuperCalculus View Post
    So yeah, I'm looking at British Mathematical Olympiad questions, and this one has me utterly stumped. Any ideas?

    Find all integers x, y and z such that:

    x^2 + y^2 + z^2 = 2(yz + 1)
    And
    x + y + z = 4018.
    Write the first equation as x^2 + (y-z)^2 = 2. Then clearly both x and yz must be 1 or 1. Is that enough of a hint?
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    "Then clearly both x and y–z must be 1 or –1."

    How the hell did you get that? O_o
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    Quote Originally Posted by satx View Post
    "Then clearly both x and y–z must be 1 or –1."

    How the hell did you get that? O_o
    What other integers could make 2 when squared and added?
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    Oh, my bad. Didn't see the "integers" part.
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  6. #6
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by Opalg View Post
    Write the first equation as x^2 + (y-z)^2 = 2. Then clearly both x and yz must be 1 or 1. Is that enough of a hint?
    Yeah, I reckon I've got it providing that is true, but can you tell me how you got to that equation? (I understand how it can only be 1 or -1)
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  7. #7
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    Quote Originally Posted by SuperCalculus View Post
    Quote Originally Posted by Opalg View Post
    Write the first equation as x^2 + (y-z)^2 = 2. Then clearly both x and yz must be 1 or 1. Is that enough of a hint?
    Yeah, I reckon I've got it providing that is true, but can you tell me how you got to that equation? (I understand how it can only be 1 or -1)
    If x^2+y^2+z^2 = 2(yz+1) = 2yz+2 then x^2+y^2-2yz+z^2 =  2. If you're aiming to do olympiad questions then you ought to be able to factorise y^2-2yz+z^2.
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  8. #8
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by Opalg View Post
    If x^2+y^2+z^2 = 2(yz+1) = 2yz+2 then x^2+y^2-2yz+z^2 =  2. If you're aiming to do olympiad questions then you ought to be able to factorise y^2-2yz+z^2.
    Ugh, I knew it was something obvious. I was thinking in all these complicated terms, and all I had to do was basic rearranging >.>
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