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

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

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

• The one you found isn't even correct - the question demands positive integers – Hagen von Eitzen Jan 20 at 17:01
• You are right. I didn't know that 0 is not a positive integer. I have edited my question. – Shromi Jan 20 at 17:04
• 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? – SmileyCraft Jan 20 at 17:06
• @SmileyCraft I understood it now. Thanks for simple explanation. – 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 $$\left(x^2+\frac x2\right)^2=x^4+x^3+\frac14x^2 and $$\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$$ is odd. Then \begin{align}\left(x^2+\frac {x-1}2\right)^2&=x^4+x^3-\frac34x^2-\frac12x+\frac14\\&=y^2-\frac14(7x^2+5x+3) and \begin{align}\left(x^2+\frac {x+1}2\right)^2&=x^4+x^3+\frac54x^2+\frac12x+\frac14\\ &=y^2+\frac14(x^2-2x-3)\\&=y^2+\frac14(x+1)(x-3)\\&\ge y^2\end{align} with equality iff $$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.