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Math Help - find integer roots

  1. #1
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    find integer roots

    need some help pleasee

    thank you very much

    Prove that if p,q,r and s are odd integers, then

    x^10 + px^9 - qx^7 + rx^4 - s = 0

    has no integer roots

    thank you in advance
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  2. #2
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    Hello, ieatfood!

    Prove that if p,\,q,\,r,\,s are odd integers,
    then: x^{10} + px^9 - qx^7 + rx^4 - s \;= \;0 has no integer roots.

    [1] The sum of the ten roots is -p, an odd integer.
    [2] The product of the ten roots is s, an odd integer.

    From [2] . . . Since s is odd, all ten roots must be odd integers.

    But the sum of ten odd integers is even . . . which contradicts [1].

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    ...

    [1] The sum of the ten roots is -p, an odd integer.
    [2] The product of the ten roots is s, an odd integer.

    ...

    How do we know this again?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, ieatfood!


    [1] The sum of the ten roots is -p, an odd integer.
    [2] The product of the ten roots is s, an odd integer.

    From [2] . . . Since s is odd, all ten roots must be odd integers.

    But the sum of ten odd integers is even . . . which contradicts [1].
    So then q and r don't matter? I was wondering about these.

    -Dan
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    ... I was wondering about these.

    -Dan
    Thank God i'm not the only one! I felt stupid asking that question

    we stand together, topsquark
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    Thank God i'm not the only one! I felt stupid asking that question

    we stand together, topsquark
    Yeah, but I know where Soroban got the theorem from...

    I just don't know the name of it. (People keep telling me, but I always forget.)

    -Dan
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  7. #7
    Senior Member DivideBy0's Avatar
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    I'd be happy if one of you could provide us with some information about it

    (How awesome are these Skype smilies!)
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  8. #8
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    In the expansion of \prod\limits_{k = 1}^N {\left( {x - r_k } \right)} = 0 the coefficient of x^{N - k} ,\;k = 1,2, \cdots N is \left( { - 1} \right)^k S_k .
    Where S_k is the sum of the products of the roots taken k at a time.
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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, ieatfood!


    [1] The sum of the ten roots is -p, an odd integer.
    [2] The product of the ten roots is s, an odd integer.

    From [2] . . . Since s is odd, all ten roots must be odd integers.
    And why must there be 10 real roots?

    As far as I can see there are at most 6 real roots (but there must be an even number of them).

    RonL
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  10. #10
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    hey guys

    im really desperate with this question..and i still dont understand it....PLEASE help!!!!! how do you know that there are 10 roots??im confused

    thank you again
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  11. #11
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    If you take the polynomial f(x) = (x-x_0)(x-x_1)...(x-x_n) and multiply them out you get as coefficients (without signs) the so-called "symettric functions":
    x_0+x_1+...+x_n
    x_0x_1+x_0x_2+...+x_{n-1}x_n (all possible products)
    ...
    x_0x_1...x_n

    The important thing about these is that they are invariant under and premutation of the roots.

    This is the (unsigned) version of the theorem Soroban used.
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