# $y^2 = \frac{x^5 - 1}{x-1}$ & $x,y \in \mathbb{Z}$ [closed]

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.

## closed as unclear what you're asking by Servaes, Daniel W. Farlow, martini, Henrik, Adam HughesDec 21 '16 at 0:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• It is indeed the same question. What part of the answer is not clear to you? – Servaes Dec 20 '16 at 20:08
• @Servaes How do we get (x,y) = (3, 11) ?? And why he assume y^2 = 0 or 1 :( – Rezwan Arefin Dec 20 '16 at 20:10
• If you don't want this to be marked as a duplicate, you should state in what sense your question is different: Is there something particular in the answer you didn't understand? Would you like to see some intuition behind the idea? Do you want a complete solution? Note: there's nothing wrong with asking for such clarification (on the contrary!) but it has to be clear what you want to know. – punctured dusk Dec 20 '16 at 20:10
• Both these statements are not part of the answer; have you read the answer? – Servaes Dec 20 '16 at 20:11
• Note that you could also ask the person who wrote that answer to edit and give more explanation, which in this case seems more appropriate IMO. I.e. it's not always necessary to ask a new question. (By the way, well done to find that question and link to it!) – punctured dusk Dec 20 '16 at 20:15

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.