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Math Help - Sequences and primes

  1. #1
    Super Member PaulRS's Avatar
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    Sequences and primes

    Consider an integer sequence {\left\{ {{a_n}} \right\}_{n \in \mathbb{N}}}<br />
satisfying: {a_n} = a \cdot {a_{n - 1}} + b \cdot {a_{n - 2}} for n \geqslant 2 ( a,b \in \mathbb{Z} )

    Let p be a prime, p \leqslant \tfrac{n}<br />
{2} (when this is possible) . Prove that {a_n} \equiv a \cdot {a_{n - p}} + b \cdot {a_{n - 2p}}\left( {\bmod .p} \right)

    Have fun!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by PaulRS View Post
    Consider an integer sequence {\left\{ {{a_n}} \right\}_{n \in \mathbb{N}}}<br />
satisfying: {a_n} = a \cdot {a_{n - 1}} + b \cdot {a_{n - 2}} for n \geqslant 2 ( a,b \in \mathbb{Z} )

    Let p be a prime, p \leqslant \tfrac{n}<br />
{2} (when this is possible) . Prove that {a_n} \equiv a \cdot {a_{n - p}} + b \cdot {a_{n - 2p}}\left( {\bmod .p} \right)

    Have fun!
    the main idea is to prove that for any m \leq n/2 we have a_n=\sum_{j=0}^m \binom{m}{j}a^{m-j}b^j a_{n-m-j}. the proof is by induction over m and noting that if m+1 \leq n/2, then:

    \sum_{j=0}^{m+1} \binom{m+1}{j}a^{m+1-j}b^ja_{n-m-j-1}=\sum_{j=0}^{m+1} \left[\binom{m}{j-1} + \binom{m}{j} \right]a^{m+1-j}b^ja_{n-m-j-1}

    =\sum_{j=0}^m \binom{m}{j}a^{m-j}b^{j+1}a_{n-m-j-2} + \sum_{j=0}^m \binom{m}{j}a^{m+1-j}b^j a_{n-m-j-1}

    =\sum_{j=0}^m \binom{m}{j}a^{m-j}b^j(aa_{n-m-j-1} + ba_{n-m-j-2})=\sum_{j=0}^m \binom{m}{j}a^{m-j}b^ja_{n-m-j}=a_n, by induction hypothesis.

    if m=p \leq n/2 is a prime, then \binom{p}{j} \equiv 0 \mod p, for all 0 < j < p. so the above gives us: a_n \equiv a^p a_{n-p} + b^p a_{n-2p} \mod p. but for any integer c: \ c^p \equiv c \mod p. Q.E.D.
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