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Math Help - NIce one, who will do it first??

  1. #1
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    NIce one, who will do it first??

    \left\{ \begin{array}{l}a^{5} = 5b^{3} - 4c \\<br />
b^{5} = 5c^{3} - 4a \\<br />
c^{5} = 5a^{3} - 4b\end{array} \right\}
    Last edited by CaptainBlack; September 1st 2007 at 01:00 PM. Reason: to correct LaTeX
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  2. #2
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    Hello, Ununuquantium!

    Didn't you Preview your work before you posted it?


    \begin{Bmatrix}a^5 \:=\: 5b^3 - 4c \\<br />
b^5 \:= \:5c^3 - 4a \\<br />
c^5 \:= \:5a^3 - 4b\end{Bmatrix}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Ununuquantium View Post
    \left\{ \begin{array}{l}a^{5} = 5b^{3} - 4c \\<br />
b^{5} = 5c^{3} - 4a \\<br />
c^{5} = 5a^{3} - 4b\end{array} \right\}
    If you are setting problems to challenge people put them in one of the Lounge groups.
    These groups are for people who want help with stuff they don't know the answer to.

    RonL
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  4. #4
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    This problem is still unsolved, maybe someone will help?
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  5. #5
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    Maybe it will help?

    If you add these equations you get:

    (a-2)(a-1)a(a+1)(a+2) + (b-2)(b-1)b(b+1)(b-2) + (c-2)(c-1)c(c+1)(d+1) = 0

    One solution is: a=b=c=2 or 1 or 0 or -1 or -2

    I don't know if there are other solutions, maybe someone can proof that a,b,c must be equal?
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  6. #6
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    Yep, I know the same solution, but I cannot prove that there are not any other solution.....
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jazzpiano View Post
    If you add these equations you get:

    (a-2)(a-1)a(a+1)(a+2) + (b-2)(b-1)b(b+1)(b-2) + (c-2)(c-1)c(c+1)(d+1) = 0

    One solution is: a=b=c=2 or 1 or 0 or -1 or -2

    I don't know if there are other solutions, maybe someone can proof that a,b,c must be equal?
    Clever!

    How about any combination of a = 0, c = -b where b is any real number.

    -Dan
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  8. #8
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    Argh..

    Quote Originally Posted by topsquark View Post
    Clever!

    How about any combination of a = 0, c = -b where b is any real number.

    -Dan
    I'm afraid it won't work...

    Despite the fact, that in equation I wrote (a-2)(a-1).... you can pick eg. a=1 b=2 c=-1 and it seems to be correct,when you put that numbers to original equations it's incorrect...
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