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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.