Find $$(x,y)$$

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jayon
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Joined: Thu Dec 08, 2016 12:10 am

Find $$(x,y)$$

Unread post by jayon » Sat Dec 17, 2016 10:00 pm

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super boy
Posts: 5
Joined: Tue Dec 20, 2016 1:11 pm

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

Unread post by 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.

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jayon
Posts: 20
Joined: Thu Dec 08, 2016 12:10 am

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

Unread post by 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...

soyeb pervez jim
Posts: 2
Joined: Sat Jan 28, 2017 11:06 pm

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

Unread post by 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 $

Driggers
Posts: 1
Joined: Tue Aug 29, 2017 6:05 pm

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

Unread post by Driggers » Tue Sep 05, 2017 1:20 pm

hey Super Boy that's a smart way of doing it.

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