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I need some help on solving this problem:

Find all integer solutions for this following equation:

$1+x+x^2+x^3=y^2$

My attempt:

Clearly $y^2 = (1+x)(1+x^2)$, assuming the GCD[$(1+x), (1+x^2)] = d$, then if $d>1$, $d$ has to be power of 2. This implies that I can assume: $1+x=2^s*a^2, 1+x^2=2^t*b^2$. If $t=0$ then it is easy to finish. Considering $t>0$, we can get $t=1$ (simple steps only), so I come up with a "Pell-related" equation .. Then I get sticking there. It has a solution $x=7$, so I guess it's not easy to find the rest.

Please help.

Nguyễn Hoài Nam
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    I rewrote this into: $y^2 = (1+x)(1+x^2)$, assuming the GCD of $(1+x), (1+x^2)$ is $d$ then $d$ has to be power of 2. This implies that I can assume: $1+x=2^sa^2, 1+x^2=2^tb^2$. If $t=0$ then it is easy to finish. Considering $t>0$, we can get $t=1$ (simple steps only), so I come up with a "Pell-related" equation .. Then I get sticking there. It has a solution $x=7$, so I guess it's not easy to find the rest. – Nguyễn Hoài Nam Jul 22 '14 at 09:10
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    This could help: $1 + x + x^2 + x^3 = \frac{x^4 - 1}{x - 1}$ – Darth Geek Jul 22 '14 at 09:10
  • Then maybe you could solve the Pell-related equation $y^2 = k(1+x^2)$ and then for every $k$ select the $(x,y)$ solution such that $k = 1 + x$ – Darth Geek Jul 22 '14 at 09:14
  • I edited the question. The Pell related equation I have is: $x^2+1=2a^2$. @DarthGeek I think the Pell-related equation you mention will define the $(x,y)$ as well, and to do the next step it will not be easier. But I'll try to write down some thought :) – Nguyễn Hoài Nam Jul 22 '14 at 09:15
  • For that Pell equation, you can use use unit factorization domain to solve it. Do you know any advanced number theory? – DeepSea Jul 22 '14 at 09:22
  • Actually I have/know the formula for it's solution: gives $a_n$ the sequence which is defined as follow: $a_0=1, a_1=1, a_{n+2}=2a_{n+1}+a_n$ for all $n=2,3,...$, then the "$x$" solutions will be: $x_n=a_{2n+1}$. But this does not finish the solution, because we have to consider the value of $x+1$ too. – Nguyễn Hoài Nam Jul 22 '14 at 09:25
  • You can assume $d=1$ or $d=2$, since $1+x^2$ is never divisible by a higher power of $2$. Hint: Consider $1+x^2$ modulo $4$. – vuur Jul 22 '14 at 11:59
  • i think I've done with $d$ already. What I'm trying to so is to use the solutions of that Pell-related equation mentioned above to apply to the main one. – Nguyễn Hoài Nam Jul 22 '14 at 12:17

3 Answers3

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There is quite a bit about this equation in Dickson's History of the Theory of Numbers. Volume 1, page 56, it says Gerono, Nouv Ann Math (2) 16 (1877) 230-234 proved the only solutions are $$(x,y)=(-1,0),\quad(0,\pm1),\quad(1,\pm2),\quad(7,\pm20)$$ On page 57, Dickson references a proof by Genocchi, Nouv Ann Math (3) 2 (1883) 306-310. Lucas, Nouv Corresp Math 2 (1876) 87-88, had noted that the problem is equivalent to solving $1+x=2u^2$, $1+x^2=2v^2$, and then letting $y=2uv$. Dickson then discusses that system in Volume 2, pages 487-488. Several references are given there.

Gerry Myerson
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    Thanks for mentioning a good book including more detailed explanation about this equation! – Halk Feb 28 '24 at 20:52
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The solution is given in Ribenboim's book on Catalan's conjecture, where all Diophantine equations $$y^2=1+x+x^2+\cdots +x^k$$ are studied. For $k=3$, only $x=1$ and $x=7$ are possible.

Dietrich Burde
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  • Can you please give me more detail of the article/chapter? Unfortunately I could not find any source about the equation that is available online, or even a hard-copy of that book. – Nguyễn Hoài Nam Jul 22 '14 at 13:24
  • Then http://math.stackexchange.com/questions/442622/solutions-to-p1-2n2-and-p21-2m2-in-natural-numbers/444958#444958 will be helpful. It is connected to Pell's equation and Phytagorean triples. – Dietrich Burde Jul 22 '14 at 13:50
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This problem, and its “inverse” (q.v. Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$) were posed by Fermat in 1657 (see, for example, Mahoney pg. 337). Both lead to Pell equations (q.v. Solutions to $p+1=2n^2$ and $p^2+1=2m^2$ in Natural numbers.), which was exactly what Fermat was trying to get his contemporaries to study with him.

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Kieren MacMillan
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