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Thread: No integer solutions to x^10 + px^9 - qx^7 + rx^4 - s = 0

  1. #1
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    No integer solutions to x^10 + px^9 - qx^7 + rx^4 - s = 0

    Prove that if p,q,r and s are odd integers, then this equation has no integer roots:

    x^10 + px^9 - qx^7 + rx^4 - s = 0
    Last edited by mr fantastic; May 11th 2010 at 02:50 AM. Reason: Re-titled.
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  2. #2
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    Hello, the undertaker!

    I baby-talked my way through it . . .


    Prove that if $\displaystyle p,q,r, s$ are odd integers,

    then: .$\displaystyle x^{10} + px^9 - qx^7 + rx^4 - s \:=\: 0$ .has no integer roots.

    Note that:

    An even integer raised to any positive integral power is even: .$\displaystyle \text{(even)}^n \:=\:\text{even}$

    An odd integer raised to any positive integral power is odd: .$\displaystyle \text{(odd)}^n \:=\:\text{odd}$


    We have: .$\displaystyle \bigg[x^{10} + px^9 + rx^4\bigg] - \bigg[qx^7 + s\bigg] \;=\;0$


    Suppose $\displaystyle x$ is even.

    We have: .$\displaystyle \bigg[(even)^{10} + p(even)^9 + r(even)^4\bigg] - \bigg[q(even)^7 + (odd)\bigg] \;=\;0 $

    . . . . . . . . . $\displaystyle \underbrace{\bigg[(even) + (even) + (even)\bigg]}_{\text{(even)}} - \underbrace{\bigg[(even) + (odd)\bigg]}_{\text{(odd)}} \;=\;0 $

    And the difference of an even number and an odd number cannot be zero.



    Suppose $\displaystyle x$ is odd.

    We have: .$\displaystyle \bigg[(odd)^{10} + p(odd)^9 + r(odd)^4\bigg] - \bigg[r(odd)^4 + (odd)\bigg] \;=\;0 $

    . . . . . . . . . $\displaystyle \underbrace{\bigg[(odd) + (odd) + (odd)\bigg]}_{\text{(odd)}} - \underbrace{\bigg[(odd) + (odd)\bigg]}_{\text{(even)}} \;=\;0$

    And the difference of an odd number and an even number cannot be zero.

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  3. #3
    Super Member Bacterius's Avatar
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    Geez, that was awesome Soroban. Such a simple technique for such a seemingly intractable problem ! Too bad I can't triple-thank you for this one.

    The only improvement I can suggest is putting the "even" and "odd" tags between \mathrm{}, like $\displaystyle \mathrm{even}$.

    Otherwise,
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