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Thread: Tripos question 1903

  1. #1
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    Tripos question 1903

    x^6+x^5-2x^4-x^3+x^2+1=0
    Y=x^4-x^2+x-1

    show y satisfies a quadratic equation with rational coefficients.

    I've tried and have not be able to do so, so far.

    Do it, please.

    Thanks.
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  2. #2
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    Quote Originally Posted by berachia View Post
    $\displaystyle x^6+x^5-2x^4-x^3+x^2+1=0$
    $\displaystyle y=x^4-x^2+x-1$

    show y satisfies a quadratic equation with rational coefficients.
    I think this is just an exercise in division of polynomials. Square y, then divide by the expression that is given to be zero:

    . . . . .$\displaystyle \begin{aligned}y^2 &= (x^4-x^2+x-1)^2 \\
    &= x^8 - 2x^6 + 2x^5 - x^4 - 2x^3 + 3x^2 - 2x + 1 \\
    &= (x^2 - x + 1)(x^6 + x^5-2x^4-x^3+x^2+1) - x^4 + x^2 - x \\
    &= -y - 1\end{aligned}$

    Therefore y satisfies the quadratic equation $\displaystyle y^2+y+1=0$.

    (I'll leave you to convince yourself that this is an occasion where division by zero is justified!)
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