How many positive integer solution pairs $(x, y)$ are there for the equation $$y^2 = \frac{(x^5 - 1)}{(x-1)},\;\; x\neq 1\;?$$

Source : Bangladesh Math Olympiad 2016 Junior Catagory

I am unable to find if there are any solutions or not.

  • 1
    $\begingroup$ The one you found isn't even correct - the question demands positive integers $\endgroup$ – Hagen von Eitzen Jan 20 at 17:01
  • $\begingroup$ You are right. I didn't know that 0 is not a positive integer. I have edited my question. $\endgroup$ – Shromi Jan 20 at 17:04
  • 2
    $\begingroup$ Have you tried to find the first few values of $(x^5-1)/(x-1)=1+x+x^2+x^3+x^4$ yet? In particular, have you looked at their prime factors yet? $\endgroup$ – SmileyCraft Jan 20 at 17:06
  • $\begingroup$ @SmileyCraft I understood it now. Thanks for simple explanation. $\endgroup$ – Shromi Jan 20 at 17:20

Assume $y^2=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ for positive integers $x,y$ (and $x\ne 1$).

Case 1: $x$ is even. Then $$\left(x^2+\frac x2\right)^2=x^4+x^3+\frac14x^2<y^2$$ and $$\left(x^2+\frac x2+1\right)^2=x^4+x^3+\frac94 x^2+x+1>y^2,$$ contradiction as $y^2$ cannot be strictly between two consecutive squares..

Case 2: $x$ is odd. Then $$\begin{align}\left(x^2+\frac {x-1}2\right)^2&=x^4+x^3-\frac34x^2-\frac12x+\frac14\\&=y^2-\frac14(7x^2+5x+3)<y^2\end{align}$$ and $$\begin{align}\left(x^2+\frac {x+1}2\right)^2&=x^4+x^3+\frac54x^2+\frac12x+\frac14\\ &=y^2+\frac14(x^2-2x-3)\\&=y^2+\frac14(x+1)(x-3)\\&\ge y^2\end{align}$$ with equality iff $x=3$. In other words, we arrive at a contradiction for odd $x>3$ and at the same time see that $x=3$ leads to a solution.


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