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Thread: Tricky induction proof

  1. #1
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    Tricky induction proof

    I just can't see to figure this out.

    Consider function $\displaystyle h$ as follows:

    $\displaystyle \bullet \ h(1) = h(2) = 1$
    $\displaystyle \bullet \ h(n) = h(n-1)^2 + h(n-2)\mbox{ for }n > 2$

    Prove by induction that for all $\displaystyle n > 5$ that $\displaystyle h(n)$ and $\displaystyle h(n-1)$ are relatively prime.
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  2. #2
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    Quote Originally Posted by VENI View Post
    I just can't see to figure this out.

    Consider function $\displaystyle h$ as follows:

    $\displaystyle \bullet \ h(1) = h(2) = 1$
    $\displaystyle \bullet \ h(n) = h(n-1)^2 + h(n-2)\mbox{ for }n > 2$

    Prove by induction that for all $\displaystyle n > 5$ that $\displaystyle h(n)$ and $\displaystyle h(n-1)$ are relatively prime.
    If give you a hint to get you started. Say that $\displaystyle h(n)$ and $\displaystyle h(n-1)^2$ where not relatively prime then there is a prime $\displaystyle p$ so that $\displaystyle p$ divides $\displaystyle h(n)\text{ and }h(n-1)$. However, $\displaystyle h(n) - h(n-1)^2=h(n-2)$ and so $\displaystyle p$ would divide $\displaystyle h(n-2)$. Which means that $\displaystyle p$ a common factor for $\displaystyle h(n-1)\text{ and }h(n-2)$. Thus, we have shown is that if $\displaystyle h(n)\text{ and }h(n+1)$ are relatively prime then $\displaystyle h(n+1)\text{ and }h(n+2)$ must be relatively prime.
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