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I'm studying Elliptic Curves and EDS (Elliptic divisibility sequences) and working on Silvermans exercises 3.34 in "The arithmetic of elliptic curves":

"An EDS over $K$ is a Sequence $(W_n)_{n\geq 1}$ defined by four initial conditions $W_1, W_2, W_3, W_4 \in K$ ans satisfying the recurrence:

$W_{m+n}W_{m-n}W_1^2=W_{m+1}W_{m-1}W_n^2-W_{n+1}W_{n-1}W_m^2$

I already proved that the even terms of the Fibonacci-sequence satisfy this recurrence.

The next exercise is to generalize this result and find a subsequence of the sequence $(L_n)_{n\geq 1}$:

$L_1=1 \quad L_2=P \qquad \qquad \qquad L_{n+2}= PL_{n+1} - L_n$

which also saitsfies the recurrence above.

I already recognized the the given formula describes a Lucas-sequence : $L_n= \frac{a^n-b^n}{a-b} $ with $P=a+b, Q=ab=1$

I just need an idea to proof that such a sequence satisfies the recurrence relation above.

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  • $\begingroup$ $L_n = \alpha^n+ \beta^n$ for some $\alpha,\beta$ $\endgroup$ – reuns Jul 25 '17 at 13:20
  • $\begingroup$ Please clarify, are you seeking a closed form expreseion for $L_n$? $\endgroup$ – Cye Waldman Jul 25 '17 at 16:09
  • $\begingroup$ No, I already have the closed form $L_n=\frac{a^n-b^n}{a-b}$ with $P=a+b, Q=ab$ . Just need an Idea to proof that such Lucas-sequence with $Q=1$ staisfies the elliptic recurrence relation. $\endgroup$ – mainman Jul 25 '17 at 17:41

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