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A few days ago I saw a question about the diophantine equation $p^2+p+1=q^\alpha$ in What is prime power of this equation of p?

and later in

A Diophantine equation with prime powers

I want the similar results about the following diophantine equation $p^2-p+1=q^\alpha$ and $p^2-p+1=3q^\alpha$, where $p$ and $q$ are prime numbers.

Of course we know that $19^2-19+1=7^3$, but is there any answer for these equations? In fact in very special cases we see $\alpha=1$ and so is there only finite answers where $\alpha=1$?

As I checked for p<400000 only there exists one solution as above. Any comments and hints are highly appreciated.

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  • $\begingroup$ You have both $a$ and $\alpha$, and may want to fix it. Also, have your read the comments to Geoff Robinson's answer here: mathoverflow.net/a/206941/16537? $\endgroup$
    – Salvo Tringali
    May 25, 2015 at 6:38
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    $\begingroup$ The post mathoverflow.net/questions/207024/… refers to a paper of Nagell in which the equations $x^2+x+1=y^n$ and $x^2+x+1=3 y^n$ for $n \ge 3$. You want to take $x=-p$, $y=q$, $n=\alpha$ which reduces you to the case where $\alpha=1$ or $2$. $\endgroup$
    – Siksek
    May 25, 2015 at 8:40
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    $\begingroup$ Nowadays such equations are treated using the primitive divisor theorem of Bilu, Hanrot and Voutier. See for example page 420 of "Number Theory: Volume I: Tools and Diophantine Equations" by Henri Cohen. $\endgroup$
    – Siksek
    May 25, 2015 at 8:45
  • $\begingroup$ Thank you very much for the useful comments. For the first equation when $n=3^b$, can we discuss about the solutions? $\endgroup$
    – BHZ
    May 25, 2015 at 9:25
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    $\begingroup$ The equation $x^2+x+1=y^3$ is an elliptic curve with trivial Mordell--Weil group. It has no integral points. $\endgroup$
    – Siksek
    May 28, 2015 at 22:56

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Nagell proved in 1921 that the equations $$p^2\pm p+1=q^\alpha$$ and $$p^2\pm p+1=3q^\alpha$$ have no nontrivial integral solutions for odd $\alpha$ except possibly for the first equation when $\alpha=3^k$ for some positive integer $k$.

The equation $p^2\pm p+1=q^2$ has no nontrivial integer solutions, while the equation $p^2\pm p+1=q^3$ have no nontrivial integer solutions except $q^\alpha=7^3=343$, giving $p=18$ (not prime) to $$p^2+p+1=q^3$$ and $p=19$ (prime) to $$p^2-p+1=q^3$$

Hence $p^2\pm p+1=3q^2$ for some prime power $q$.

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