shobber wrote: prove that there exists a nonnegative integer $m$ such that $n=3^m$. You mean $a_n=3^m$, right?

Anybody can help us proving this so beatiful one?

hint

Ok! We know, and then...?

We find out the general of this sequence: $ a_n = \frac {1}{2}({\frac {\sqrt {5} + 1}{2})^{2n} + (\frac {\sqrt{5} - 1}{2})^{2n}}$ I think has a problem with your post . Example $ a_2 = 7$

Can you show TTsphn, how had you received this explicite form of recurrentia and why it is the only one?

TTsphn wrote: We find out the general of this sequence: $ a_n = \frac {1}{2}({\frac {\sqrt {5} + 1}{2})^{2n} + (\frac {\sqrt {5} - 1}{2})^{2n}}$ I think has a problem with your post . Example $ a_2 = 7$ I'm sorry, TTsphn, but this formula is wrong. Just look at $ n=1$ .... . And in fact, it does not exist formulas with the form $ \alpha u^n+\beta v^n$ for such sequence.

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$ a_n=L_{2n}$.but continue....

I think i 've just figured an ugly solution for this magnificient problem: As shimizu-nin said we have:$ a_n=L_{2n}=L_n^2-2(-1)^n$ So:$ a_n+(-1)^n=L_n^2-(-1)^n=p.$ It's easy to show that: $ n$ must be odd.Firts we note that:$ F_3=2 ; L_3=4$ and $ (F_m,F_n)=F_{(m,n)};(L_m,L_n)=L_{(m,n)}$ (for both $ m,n$ are odd only).By some expensive step we obtain: $ 5F_n^2+3L_n^2=4p(1)$ Decomposing that:$ n$ isnot divisible by $ 3$ we yields that the both $ F_n$ and $ L_n$ are odd so the $ LHS$ is divisible by $ 8$=>Contradiction! So we have :$ n=3m$.Now we descript $ (1)$ through $ F_m$ and $ L_m$ .I think we must have again this divisibility! (check it!)

This's not ugly solution, welldone

pluricomplex wrote: This's not ugly solution, welldone I consider it as ugly one because I really dont want to check the last step.But I believe it's true!

of cause it's trivial true, just note that $ n$ is odd and $ 3\mid n$ so $ 4\mid L_n$, that's all we need

shobber wrote: Sequence $\{ a_n \}$ satisfies: $a_1=3$, $a_2=7$, $a_n^2+5=a_{n-1}a_{n+1}$, $n \geq 2$. If $a_n+(-1)^n$ is prime, prove that there exists a nonnegative integer $m$ such that $n=3^m$. First: Easy to see that: $a_{1}=3,a_{2}=7, a_{n+1}=3a_{n}-a_{n-1}$ and $a_{n}=u^{n}+(\frac{1}{u})^{n}$ With $u=\frac{3+\sqrt{5}}{2}$ Let $b_{n}=a_{n}+(-1)^{n}$ we are going to prove that: Let k is max positive integer with $3^{k}\mid n $ then $b_{3^{k}}\mid b_{n}$ (*) Then with (*) we must have: if$ b_{n}$ is prime then $n=3^{k}$ Now we prove (*): Without loss of generality we do with n is odd ( if n is even that only change with modulo 3) $b_{n}=u^{n}+u^{-n}-1$ Let $x=u^{3^{k}}$ then $b^{n}=x^{r}-1+x^{-r}$ and $3^{k}r=n$, 3 not divide $r$ . We will prove $(x+1/x-1)\mid (x^{r}-1+x^{-r})$ with r is odd and not divide by 3. Without loss of generality we only do with $r=6t+1$ ($r=6t-1$ similar)(**) and (**) true with inductive: $y_{t}=x^{6t+1}-1+x^{-(6t+1)}$ and $y_{t}-y_{t-1}=(x-1+1/x)(x+1+1/x)(x^{3t-1}+x^{-(3t-1)})((x^{3k}+x^{-3k})-(x^{3k-2}+x^{-3k+2}))$ and $x^{k}+x^{-k}$ all are integer. We are done

Solution of vntbqpqh234 so buffalo(anh phúc phải ko)