My question is this:
$y^2 = \frac{x^5 - 1}{x-1}$ & $x,y \in \mathbb{Z}$
Source: BdMO 2016 Regionals Set 1 Question 3.
An equation in $\mathbb{Z}$ I think this is the exact same question. Actually I didn't get the solution in the link. Please don't mark as duplicate.
I tried solving this same problem by writing code; there seems to be $6$ solution pairs to this equation. How do I get the $6$ solutions mathematically? I indeed need a full solution. Please help.
To add some detail to the answer in the linked question, first find the roots $(0,\pm1)$ by test. Then we can assume $x \neq 0$. Multiply by $4$ to get $$(2y)^2=4x^4+4x^3+4x^2+4x+4$$ and note that $$ (2x^2+x)^2 = 4x^4+4x^3+x^2, $$ $$ (2x^2+x+1)^2 = 4x^4+4x^3+5x^2+2x+1 $$ are successive squares. Then $$(2y^2)-(2x^2+x)^2=3x^2+4x+4=2x^2+(x+2)^2 \gt 0\\ (2y)^2-(2x^2+x+1)^2=-x^2+2x+3=-(x-1)^2+4$$ The second will be less than $0$ unless $x$ is close to $1$. In that case our expression $4x^4+4x^3+4x^2+4x+4$ cannot be a square because it is between successive squares. We can have $-(x-1)+4=0$ for $x=-1,3$. Plugging in, we find that $2y$ is even and we find the other roots. You should also check $x$ values between $-1$ and $3$, which might correspond to $(2y)^2$ being a perfect square larger than $(2x^2+x+1)^2$, but none work.