Find $$(x,y)$$

[jayon]

....................

[super boy]

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.

[jayon]

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]

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]

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