Re: um.....this is...!

From: Ignacio Larrosa Cañestro (ilarrosaQUITARMAYUSCULAS_at_mundo-r.com)
Date: 10/06/04 

Date: Wed, 6 Oct 2004 14:02:03 +0200

"mina_world" <mina_world@hanmail.net> escribió en el mensaje
news:cjvfcl$gc0$1@news.hananet.net...
>
> "Ignacio Larrosa Ca?stro" <ilarrosaQUITARMAYUSCULAS@mundo-r.com> wrote in
> message news:2sfst4F1kkvtbU1@uni-berlin.de...
>> En el mensaje:cjudam$qor$1@news.hananet.net,
>> mina_world <mina_world@hanmail.net> escribi?
>> > hello.....doctor~
>> >
>> > a_1 = 1
>> >
>> > 2*{a_(n+1)} = 3*a_n + sqrt{(5(a_n)^2) + 4} (n >= 1)
>> >
>> > show that there does not exist n such that 1989 | a_(2n).
>> > (n is positive integer)
>> >
>> > ----------------------------------------------------
>> >
>> > um....it's so difficult to me.
>> >
>> > i need your advice.
>> >
>> > thank you very much for your advice.
>>
>> Hint: Prove that a(n) = Fibonacci(2n)
>>
>>
>
> um.....
> a_1 = 0
> a_2 = 1
> a_3 = 3
> a_4 = 8
> a_5 = 21
> a_6 = 55
No,

a(1) = 1 (your definition!)
a(2) = 3
a(3) = 8
a(4) = 21
...

> a_(n+1) - 3.a_n + a_(n-1) = 0 by rule.

Ok, writing

2a(n+1) - 3a(n) = sqrt(5a(n)^2 + 4)

4a(n+1)^2 + 9a(n)^2 - 12a(n+1)a(n) = 5a(n)^2 + 4 ===>

4a(n+1)^2 + 4a(n)^2 - 12a(n+1)a(n) = 4 ===>

a(n+1)^2 + a(n)^2 - 3a(n+1)a(n) = 1 ===>

replacing n with n-1,

a(n)^2 + a(n-1)^2 - 3a(n)a(n-1) = 1 ===>

substrctinge that last equations,

a(n+1)^2 - a(n-1)^2 - 3a(n)(a(n+1)- a(n-1)) = 0

a(n+1) + a(n - 1) - 3a(n) = 0

(because a(n+1) =/= a(n-1) !)

Its says,

a(n+1) = 3a(n) - a(n-1), a(1) = 1

The Fibonacci sequence is

F(1) = F(2) = 1, F(n+1) = F(n) + F(n-1)

Then F(n+2) = F(n+1) + F(n) = 2F(n) + F(n-1)

F(n+3) = F(n+2) + F(n+1) = 3F(n) + 2F(n-1)

F(n+4) = F(n+3) + F(n+2) = 5F(n) + 3F(n-1) = 3(2F(n) + F(n-1)) - F(n) =
3F(n+2) - F(n)

Then, the sequence b(n) = F(2n) verify

b(n+2) = F(2n + 4) = 3F(2n + 2) - F(2n) = 3b(n+1) - b(n)

b(1) = F(2) = 1

b(2) = F(4) = 3

Then b(n) and a(n) verify the same requrrence relation with the same initial
values, then are equal.

Is says,

a(n) = F(2n)

It is easy to prove by induction that k | F(n) ===> k | F(mn)

As 1989 = 3^2*13*17 and

F(7) = 13 = 0 (mod 13)

F(9) = 34 = 0 (mod 17)

F(12) = 144 = 0 (mod 9)

F(LCM(7, 9, 12)) = F(252) = 0 (mod 9*13*17)

Then a(126) = 0 (mod 1989) and 1989 | a(2*63) 

-- 
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
>
> a_n = (1/sqrt(5))*{((3+sqrt(5))/2)^(n-1)  -  ((3-sqrt(5))/2)^(n-1)}
> = (1/sqrt(5))*(2/(3+sqrt(5)))*{((3+sqrt(5))/2)^n  - ((3-sqrt(5))/2)^n}
> by characteristic equation.
>
> and
>
> a_2n = (1/sqrt(5))*(2/(3+sqrt(5)))*{((3+sqrt(5))/2)^(2n)  -
> ((3-sqrt(5))/2)^(2n)}
> = (1/sqrt(5))*(2/(3+sqrt(5)))*{((14+6.sqrt(5))/4)^n  - 
> ((14-6.sqrt(5))/4)^n}
>
> and
>
> let a_2n = b_n
> b_(n+2) = 7.b_(n+1) - b_n by reverse pursuit to characteristic equation.
>
> and
>
> b_1 = a_2 = 1 (mod 9)
> b_2 = a_4 = 8 (mod 9)
>
> and
>
> suppose that
> b_(2n-1) = 1 (mod 9)
> b_2n = 8 (mod 9)
>
> b_(2n+1) = 7.b_(2n) - b_(2n-1)  = 7.8 - 1 = 55 = 1 (mod 9)
> b_(2n+2) = 7.b_(2n+1) - b_(2n) = 7.1 - 8 = -1 = 8 (mod 9)
>
> by mathematical induction,
> b_(2n) = 8 (mod 9)
> b_(2n+1) = 1 (mod 9)
>
> but, 1989 = 0 (mod 9)
>
> thus, 1989 | b_n = a_2n is impossible.
> -------------------------------------------------------------
>
> um.....is it insufficient ??
>
> thank you very much.
>
>

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