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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.

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Rezwan Arefin
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  • 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
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    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. – Bart Michels 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
  • @barto Yes.. Actually I wanted to have a full solution .. I didn't get the solution in the problem... I mean it is unclear to me – Rezwan Arefin Dec 20 '16 at 20:12
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    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!) – Bart Michels Dec 20 '16 at 20:15
  • @RezwanArefin Hey! I am from Bangladesh too. I posted the exact same question on the BdMO problem --- http://math.stackexchange.com/questions/2070615/formula-for-calculating-the-total-number-of-solution-pairs-to-an-equation?noredirect=1&lq=1 – Soha Farhin Pine Dec 25 '16 at 14:13
  • @SohaFarhinPine You are not allowed to post same things in stackexchange :| – Rezwan Arefin Dec 26 '16 at 07:55
  • @RezwanArefin For your information, my question isn't the same question as yours. Your question was solely based on how to solve the particular problem. But, my question isn't that. I solved the problem. My question is; is there a formula in general to find out the number of rational solutions of a Diophantine equation. My question is way more wider than yours. I solved the problem in my question, you can follow that method. – Soha Farhin Pine Dec 26 '16 at 09:58
  • See also: Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. – Martin Sleziak Jan 22 '18 at 11:57

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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.

Ross Millikan
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