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Thread: Roots of a quadratic

  1. #1
    MHF Contributor alexmahone's Avatar
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    Roots of a quadratic

    Let $\displaystyle \alpha$ and $\displaystyle \beta$ be the roots of the quadratic equation $\displaystyle x^2 + mx - 1 = 0$, where m is an odd integer. Let $\displaystyle \lambda_n = \alpha^n + \beta^n$, for $\displaystyle n \ge 0$. Prove that for $\displaystyle n \ge 0$,
    (a) $\displaystyle \lambda_n$ is an integer; and
    (b) gcd $\displaystyle (\lambda_n, \lambda_{n + 1}) = 1$.
    Last edited by alexmahone; Oct 24th 2008 at 07:53 AM.
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    Let $\displaystyle \alpha$ and $\displaystyle \beta$ be the roots of the quadratic equation $\displaystyle x^2 + mx - 1 = 0$, where m is an odd integer. Let $\displaystyle \lambda_n = \alpha^n + \beta^n$, for $\displaystyle n \ge 0$. Prove that for $\displaystyle n \ge 0$,
    (a) $\displaystyle \lambda_n$ is an integer; and
    (b) gcd $\displaystyle (\lambda_n, \lambda_{n + 1}) = 1$.
    If $\displaystyle \alpha$ and $\displaystyle \beta$ are solutions to that equation, then $\displaystyle (x-\alpha)(x-\beta)= x^2+ mx-1$. Multiplying out the left side, $\displaystyle x^2- (\alpha+ \beta)x+ \alpha\beta= mx^2+ mx- 1$

    That tells you $\displaystyle \alpha+ \beta= -m$ and $\displaystyle \alpha\beta= -1$. Does that help?
    Last edited by HallsofIvy; Oct 24th 2008 at 09:18 AM.
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