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Math Help - Tricky induction proof

  1. #1
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    Tricky induction proof

    I just can't see to figure this out.

    Consider function h as follows:

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

    Prove by induction that for all n > 5 that h(n) and 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 h as follows:

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

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