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Using completing the square and factoring method I could show that the equation $x^2+x+1=y^n$, where $x,y$ are positive odd and $n$ is positive even integers, does not have solution, but I could not show that for positive odd $x,y$ and odd $n>1$ the equation does (does not) have solution.

Thank you for your contribution.

asad
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  • Solutions where? $(0,1)$ and $(-1,1)$ clearly are solutions. – Ennar Oct 06 '16 at 06:18
  • The OP explicitly asks for $x,y,n>1$ and odd. – Giovanni De Gaetano Oct 06 '16 at 06:23
  • And also explicitly says that they can show that for $x,y$ odd and $n$ even there are no solutions. Well, $(-1,1)$ is counterexample to that, unless further restrictions are posed. Most likely that solutions should be positive integers. @GiovanniDeGaetano – Ennar Oct 06 '16 at 06:25
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    Well, $18^2 + 18 + 1 = 7^3$. – Ivan Neretin Oct 06 '16 at 06:26
  • Sorry, you wanted $x$ to be odd as well. Oops! – Ivan Neretin Oct 06 '16 at 06:35
  • I corrected the restriction. – asad Oct 06 '16 at 06:46
  • the related question $x^2 + 3 = y^n$ is done completely in An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica and Ion Cucurezeanu. The core idea is that the Eisenstein integers are a unique factorization domain. For you, $x^2 + x + 1 = (x - \omega)(x - \omega^2),$ where $\omega$ is a nontrivial cube root of unity. – Will Jagy Oct 06 '16 at 17:49
  • https://en.wikipedia.org/wiki/Eisenstein_integer – Will Jagy Oct 06 '16 at 17:53
  • Now posted also on MathOverflow: http://mathoverflow.net/questions/251633/is-x2x1-ever-a-perfect-power As usually, I will add link to this post. And there are also several other discussions of [meta-tag:cross-posting] on meta. – Martin Sleziak Oct 07 '16 at 23:55

1 Answers1

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Your equation can be rewritten as

$$\frac{x^{3}-1}{x-1} = y^{N}.$$

The Diophantine equation

$$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \tag1$$

was the subject matter of a couple of papers of T. Nagell from the $1920's.$ Some twenty-odd years later, W. Ljunggren clarified some points in Nagell’s arguments and completed the proof of the following result.

Theorem. Apart from the solutions

$$\frac{3^{5}-1}{3-1}=11^{2}, \quad \frac{7^{4}-1}{7-1}=20^{2}, \quad \frac{18^{3}-1}{18-1} = 7^{3},$$

the equation in $(1)$ has no other solution $(x, y, n, q)$ if either one of the following conditions is satisfied:

$(1.$ $q = 2$,

$(2.$ $3$ divides $n$,

$(3.$ $4$ divides $n$,

$(4.$ $q =3$ and $n$ is not congruent with $5$ modulo $6$.

Clearly enough, this theorem implies that there are only two solutions to the equation you are considering: $(x=1, y=3, N=1)$ and $(x=18, y=7, N=3)$.

user401699
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