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Home » Olympiad Section » Number Theory » Number Theory Unsolved Problems

sequence
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kiemkhach
P versus NP

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sequence
my book

consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$
and $a_{n+1}=6a_n-a_{n-1}-4$

find $n$ such that $a_n$ is a perfect square

Posted: Wed Feb 16, 2005 3:47 am

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RobertuX
Poincare Conjecture

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Re: sequence
my book

kiemkhach wrote:

consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$
and $a_{n+1}=6a_n-a_{n-1}-4$

find $n$ such that $a_n$ is a perfect square

Well, $n=1$ and $n=2$ are two solutions, rightt???? Razz

but I suppose you want an $n>2$ .
And in general

$a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).$

TCM

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Posted: Wed Feb 16, 2005 6:49 am

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kiemkhach
P versus NP

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Re: sequence
my book

RobertuX wrote:

kiemkhach wrote:

consider a sequence $(a_n)$ satisfies $a_0=1;a_1=9$
and $a_{n+1}=6a_n-a_{n-1}-4$

find $n$ such that $a_n$ is a perfect square

Well, $n=1$ and $n=2$ are two solutions, rightt???? Razz

but I suppose you want an $n>2$ .
And in general

$a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).$

TCM

$a_3=6a_2-a_1-4=54-1-4=49=7^2$ ???? Sad

can you post your solution? Mr. Green

I think ;n=1 and n=2 and n=3 are solutions??

Posted: Wed Feb 16, 2005 10:14 pm

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RobertuX
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Re: sequence
my book

RobertuX wrote:

In general
$a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).$

Well, of course this 3 values $n=1,2,3$ are solutions, and using my expression $a_4$ is NOT a square.

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Posted: Fri Feb 18, 2005 3:49 pm

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kiemkhach
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Re: sequence
my book

RobertuX wrote:

In general
$a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).$

Well, of course this 3 values $n=1,2,3$ are solutions, and using my expression $a_4$ is NOT a square.

I don't understand! Sad

In general : $a_n=1+\sqrt{2}((3+\sqrt{2})^n-(3-\sqrt{2})^n)$
how do you prove $a_n$ is not a square???( $n>3$ )

Posted: Sat Feb 19, 2005 2:30 am

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RobertuX
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Re: sequence
my book

kiemkhach wrote:

In general
$a_n=1+\sqrt 2\left((3+2\sqrt 2)^n-(3-2\sqrt 2)^n\right).$
how do you prove $a_n$ is not a square???( $n>3$ )

Well, sorry for the last comment, by definition $a_{n+1}=6a_n-a_{n-1}-4$ , this is so colled non homogeneous sequence, so first I get an equivalent homogeneous sequence, HOW? well, in this case the sum of coefficients of the homogeneous part is $1-6+1=-4$ so we need to consider $b_n=a_n-1$ , in fact $b_n$ satisfies the relations: $b_0=0$ , $b_1=8$ and $b_{n+1}-6b_n+b_{n-1}=0$ and to solve this kind of sequences you need to consider the charasteristic polynomial related to the sequence, in this case is $x^2-6x+1=0$ which roots are $\alpha=3-\sqrt 8$ and $\beta=3+\sqrt 8$ , so $b_n=A\alpha^n+B\beta^n$ , now we calculate $A$ and $B$ with the initial values, so $A+B=0$ and $\alpha A+\beta B=8$ then $A=-B=\sqrt 2$ so, you can write easily the expression of $a_n$ . If you need for info please, read a book of Discrete Mathematics.

Sorry again,and best regards,

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Posted: Sat Feb 19, 2005 9:08 am

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