PROBLEM NO.49 (B0MC-2,DAY-5)

Discussion on Bangladesh National Math Camp
User avatar
SANZEED
Posts:550
Joined:Wed Dec 28, 2011 6:45 pm
Location:Mymensingh, Bangladesh
PROBLEM NO.49 (B0MC-2,DAY-5)

Unread post by SANZEED » Wed Apr 04, 2012 11:55 pm

Let $p$ be a prime ,and let ${a_{k}}$ be an infinite sequence of integers for$k=0,1,2,......$ such that $a_{0}=0,a_{1}=1$,and $a_{k+2}=2a_{k+1}-pa_{k}$ for $k\ge 0$.if $-1$ appears in the sequence ,find all possible values of $p$.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

User avatar
SANZEED
Posts:550
Joined:Wed Dec 28, 2011 6:45 pm
Location:Mymensingh, Bangladesh

Re: PROBLEM NO.49 (B0MC-2,DAY-5)

Unread post by SANZEED » Thu Apr 05, 2012 12:08 am

Hint:
Taking mod $p,p-1$ makes it easier.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: PROBLEM NO.49 (B0MC-2,DAY-5)

Unread post by Phlembac Adib Hasan » Thu Apr 05, 2012 10:29 am

SANZEED wrote:Hint:
Taking mod $p,p-1$ makes it easier.
No need to do so.$a_2=2.1-p.0=2$.
So $a_3=2.2-p.1=4-p$
Which gives if $p>5$, it's impossible to get $-1$ in the sequence as $a_i$ will be decreasing then.Now plug in $p=2,3,5$ which gives the result $p=5$.
Also replace the part $k\ge 0$ by $k\ge 2$ in the problem.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
SANZEED
Posts:550
Joined:Wed Dec 28, 2011 6:45 pm
Location:Mymensingh, Bangladesh

Re: PROBLEM NO.49 (B0MC-2,DAY-5)

Unread post by SANZEED » Sat Apr 07, 2012 9:35 am

many many thanks Adib.my solution took 2 pages.I used fermat's little theorem too.
:idea: :|
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

Post Reply