[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include(/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php) [function.include]: failed to open stream: No such file or directory
[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include() [function.include]: Failed opening '/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php' for inclusion (include_path='.:/opt/php53/lib/php')
[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include(/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php) [function.include]: failed to open stream: No such file or directory
[phpBB Debug] PHP Warning: in file [ROOT]/includes/bbcode.php on line 122: include() [function.include]: Failed opening '/home/shoeb/public_html/www.matholympiad.org.bd/forum/includes/phpbb-latex.php' for inclusion (include_path='.:/opt/php53/lib/php')
[phpBB Debug] PHP Warning: in file [ROOT]/includes/session.php on line 1042: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4786: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4788: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4789: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
[phpBB Debug] PHP Warning: in file [ROOT]/includes/functions.php on line 4790: Cannot modify header information - headers already sent by (output started at [ROOT]/includes/functions.php:3887)
BdMO Online Forum • View topic - Find $$(x,y)$$

Find $$(x,y)$$

For students of class 9-10 (age 14-16)
Facebook Twitter

Find $$(x,y)$$

Post Number:#1  Unread postby jayon » Sat Dec 17, 2016 10:00 pm

....................
Attachments
Untitled7.png
Untitled7.png (28.3 KiB) Viewed 345 times
User avatar
jayon
 
Posts: 20
Joined: Thu Dec 08, 2016 12:10 am

Re: Find $$(x,y)$$

Post Number:#2  Unread postby super boy » Sun Dec 25, 2016 2:05 pm

The only solution is $(x,y) = (3,11) $

Solution:
First, I tried to divide $x^5 - 1$ by $x-1$. It became,

$y^2 = \frac{x^5-1}{x-1}$
$\Leftrightarrow y^2 = x^4 + x^3 + x^2 + x + 1 $

Then, I think to put an inequality and to find an easy equation so that I can find the value of $x$.

Notice that, $(x^2 + \frac{x}{2} )^2 = x^4 + x^3 + \frac{x^2}{4} $
and, $( x^2 + \frac{x}{2} + 1 )^2 = x^4 + x^3 + \frac{9x^2}{4} + x + 1$

Now, $ x^4 + x^3 + \frac{x^2}{4} < x^4 + x^3 + x^2 + x + 1 < x^4 + x^3 + \frac{9x^2}{4} + x + 1 $
$\Rightarrow (x^2 + \frac{x}{2} )^2 < y^2 < ( x^2 + \frac{x}{2} + 1 )^2 $
$\Rightarrow x^2 + \frac{x}{2} < y < x^2 + \frac{x}{2} + 1 $

If $x = 2k, k\in\mathbb{N}$, then it'll be,

$4k^2+k<y<4k^2+k+1 \Longrightarrow y\notin\mathbb{N}$ $[\because (x,y) \in\mathbb{N} ]$

So, there is no solution if $y$ is even number.

If $x=2k+1,k\in\mathbb{N}$, then it'll be,

$ 4k^2+5k+\frac{1}{2} + 1<y<4k^2+5k+1+\frac{1}{2}+1 $
$ \Rightarrow y = 4k^2+5k+2 = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2} $
$ \Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2 $
$ \Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4} $
$ \Rightarrow x^2 - 2x - 3 = 0 $
$ \Rightarrow (x - 3)(x + 1) = 0 $

So, $ x = 3 , y = 11 $


I'll really appreciate if anyone check the solution to tell me if it's correct or not.
super boy
 
Posts: 5
Joined: Tue Dec 20, 2016 1:11 pm

Re: Find $$(x,y)$$

Post Number:#3  Unread postby jayon » Wed Jan 11, 2017 8:32 pm

super boy wrote:The only solution is $(x,y) = (3,11) $

Solution:
First, I tried to divide $x^5 - 1$ by $x-1$. It became,

$y^2 = \frac{x^5-1}{x-1}$
$\Leftrightarrow y^2 = x^4 + x^3 + x^2 + x + 1 $

Then, I think to put an inequality and to find an easy equation so that I can find the value of $x$.

Notice that, $(x^2 + \frac{x}{2} )^2 = x^4 + x^3 + \frac{x^2}{4} $
and, $( x^2 + \frac{x}{2} + 1 )^2 = x^4 + x^3 + \frac{9x^2}{4} + x + 1$

Now, $ x^4 + x^3 + \frac{x^2}{4} < x^4 + x^3 + x^2 + x + 1 < x^4 + x^3 + \frac{9x^2}{4} + x + 1 $
$\Rightarrow (x^2 + \frac{x}{2} )^2 < y^2 < ( x^2 + \frac{x}{2} + 1 )^2 $
$\Rightarrow x^2 + \frac{x}{2} < y < x^2 + \frac{x}{2} + 1 $

If $x = 2k, k\in\mathbb{N}$, then it'll be,

$4k^2+k<y<4k^2+k+1 \Longrightarrow y\notin\mathbb{N}$ $[\because (x,y) \in\mathbb{N} ]$

So, there is no solution if $y$ is even number.

If $x=2k+1,k\in\mathbb{N}$, then it'll be,

$ 4k^2+5k+\frac{1}{2} + 1<y<4k^2+5k+1+\frac{1}{2}+1 $
$ \Rightarrow y = 4k^2+5k+2 $ = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2} $
$ \Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2 $
$ \Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4} $
$ \Rightarrow x^2 - 2x - 3 = 0 $
$ \Rightarrow (x - 3)(x + 1) = 0 $

So, $ x = 3 , y = 11 $


I'll really appreciate if anyone check the solution to tell me if it's correct or not.

HEI.. WHAT YOU DID WRITE IN THE LAST PART.. PLZ MAKE IT CLEAR TO UNDERSTAND...
DIDN'T YOU USE LATEX fonts..IF U DID THEN TRY TO USE IT PROPERLY DUDE...
User avatar
jayon
 
Posts: 20
Joined: Thu Dec 08, 2016 12:10 am

Re: Find $$(x,y)$$

Post Number:#4  Unread postby soyeb pervez jim » Wed May 31, 2017 1:47 am

He just forgotten to give dollar . For him I am giving that.....
The last part would be

$ y=4k^2+5k+2 = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2}
\Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2
\Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4}
\Rightarrow x^2 - 2x - 3 = 0
\Rightarrow (x - 3)(x + 1) = 0 So x = 3 , y = 11 $
soyeb pervez jim
 
Posts: 2
Joined: Sat Jan 28, 2017 11:06 pm


Share with your friends: Facebook Twitter

  • Similar topics
    Replies
    Views
    Author

Return to Secondary Level

Who is online

Users browsing this forum: Baidu [Spider] and 1 guest