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Math Help - Equation with three variables

  1. #1
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    Equation with three variables

    x^3 + y^3 + z^3 = 2008
    Find all triples (x,y,z) for this equation where x, y, and z are natural numbers

    I figured out all the triples just through trial but I don't understand how you're supposed to figure it out. Can someone help me?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    This is

    a third degree Diophantine equation...look here

    Diophantine Equation--3rd Powers -- from Wolfram MathWorld....
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  3. #3
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    Quote Originally Posted by Dslycixdued View Post
    x^3 + y^3 + z^3 = 2008
    Find all triples (x,y,z) for this equation where x, y, and z are natural numbers

    I figured out all the triples just through trial but I don't understand how you're supposed to figure it out. Can someone help me?
    The best way I see to do this is by restricting the possible cases. Say that x\geq y\geq z. Then we can see that z\leq 8 and 9 \leq x\leq 12. This gives us the some cases to try (x,y,z). And any permutation will be a solution of that.
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    Quote Originally Posted by ThePerfectHacker View Post
    Say that x\geq y\geq z. Then we can see that z\leq 8 and 9 \leq x\leq 12.
    I sort of understand that but could you explain why z\leq 8 and 9 \leq x\leq 12 ?
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  5. #5
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    Quote Originally Posted by Dslycixdued View Post
    I sort of understand that but could you explain why z\leq 8 and 9 \leq x\leq 12 ?
    Because otherwise it would be impossible to get 2008. Note, z is the smallest this means x^3+y^3+z^3\geq z^3+z^3+z^3 =3z^3 but if z\geq 9 then 3z^3 would exceede 2008. Similarly with 9\leq x. The last constraint x\leq 12 comes from because if x\geq 13 then x^3 would already exceede 2008.
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  6. #6
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    Hello,

    x>y>z

    We know that x^3+y^3+z^3=2008

    2008=x^3+y^3+z^3>z^3+z^3+z^3=3z^3

    -> z^3<2008/3

    (2008/3.0)^(1/3) ~ 8.7

    Thus z< or =8 as z is an integer.

    This goes the same way for x, but by superior values.
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