<s>[INST] Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square 
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.

My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1$ so $d\vert 2$.
Case 1: $d=1$. If $d=1$ so $(x^4+x^2+1)$,$(x+1)$ is perfect square and $(x^2+\dfrac{1}{2})^2\le x^4+x^2+1\le(x^2+1)^2$ so no satisfactory solution.
Case 2: $d=2$. If $d=2$ so $\dfrac{x^4+x^2+1}{2}$,$\dfrac{x+1}{2}$ is perfect square. 
But now I stuck. Please give me a hint. Thank you.
 [/INST] There's a simpler solution if you consider $\frac{x^6 - 1}{x - 1} = (x^2 + x + 1)(x^3 + 1)$. Let $d = \gcd(x^2 + x + 1, x^3 + 1)$, then $d \mid x^2 + x + 1 \mid x^3 - 1$ and $d \mid x^3 + 1$. So $d \mid 2$ but the number $x^2 + x + 1 = x(x + 1) + 1$ is odd. Therefore $d = 1$ and the numbers $x^2 + x + 1$ and $x^3 + 1$ are perfect squares. But for $x > 1$ we have $x^2 < x^2 + x + 1 < (x + 1)^2$. So there's no such $x$.
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