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Find all integer solutions of the equation, $n^3 = p^2 - p - 1$, where p is prime.

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    $\begingroup$ Related to elliptic curve $y^2=4n^3+5$, although even if you find a lot of rational points on that curve, restricting to $\frac{y+1}2$ being an integer prime seems hard. $\endgroup$ – Thomas Andrews Apr 23 '13 at 15:33
  • $\begingroup$ A trivial solution is $p=2$, $n=1$. If $n\ge2$, then we easily see that we must have $n<p$. As $n^3\equiv-1\pmod p$, we easily see that $n$ is of order six in $\mathbb{F}_p^*$. Consequently $p\equiv1\pmod6$. A little bit of trial and error gives another solution $n=11$, $p=37$. I don't want to go further by hand :-) $\endgroup$ – Jyrki Lahtonen Apr 23 '13 at 16:05
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    $\begingroup$ If $n\ge2$, then we have $p\mid (n^2-n+1)$ and $(n+1)\mid(p-1)$, but so what? $\endgroup$ – Jyrki Lahtonen Apr 23 '13 at 16:10
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Noting, as was essentially done previously, that $$ (8p-4)^2 = (4n)^3+80, $$ we are led to find the integral points on the elliptic curve $Y^2=X^3+80$. There are various effective methods for doing this, implemented in various computer algebra systems. Using Magma, for example, we are told that these points have $$ (X,|Y|) \in \{ (-4,4), (1,9), (4,12), (44,292) \}, $$ leading to $(p,n)=(2,3), (11,37)$.

Mordell curves like this (i.e. elliptic curves of the shape $Y^2=X^3+k$) have been completely solved (at least in terms of finding their integral points) for all $|k| < 10^7$ or so; for values up to $10^4$, there is published work of Gebel, Petho and Zimmer.

There is likely a vaguely elementary approach to this problem, as well.

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(Also a partial approach - too long for a comment.)

This problem is equivalent to showing that:

$$(2p-1)^2 - 5 = 4n^3$$

We might look at the ring of algebraic integers in $\mathbb Q[\sqrt{5}]$. This could be written as $R=\mathbb Z[\omega]$ where $\omega=\frac{-1+\sqrt{5}}{2}$.

Then we have $N(p+\omega)=N\left(\frac{2p-1 + \sqrt{5}}{2}\right)=n^3$, where $N$ is the norm. My memory is rusty, but I believe $R$ is a UFD. If $R$ is a unique factorization domain, this means that $p+\omega$ factors as a perfect cube in $R$ times some unit.

It is pretty easy to show that $p+\omega$ is not a perfect cube in $R$. But then there are the other unit cases. You can restrict yourself to the units $\omega$ and $1-\omega$, I believe. Howver, not sure how to proceed from there.

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  • $\begingroup$ Sadly $\mathbb{Q}[\sqrt{-5}]$ is not a UFD. $\endgroup$ – Captain Darling Apr 23 '13 at 20:50
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    $\begingroup$ @CaptainDarling I'm in $\mathbb Q[\sqrt{5}]$ not $\sqrt{-5}$. $\endgroup$ – Thomas Andrews Apr 23 '13 at 20:56
  • $\begingroup$ Indeed, Wikipedia says that it is then a UFD, sorry! $\endgroup$ – Captain Darling Apr 23 '13 at 21:45
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Rewrite the expression as $p(p-1)=(n+1)(n^2-n+1)$. Case 1: $p|n+1$. It follows that $p\leq n+1$. But then also $p-1 \geq n^2-n+1$, and from these we get $n \geq n^2-n+1$ which can only be true for $n=1$ which gives $p=2$.

Case 2: $p|n^2-n+1$. We write $\frac{n^2-n+1}{p}=k$. Plugging this into the equation we get $p=kn+k+1$. Return to the definition of $k$ and plug this in. We get $n^2-n+1=k(kn+k+1)$ which translates to the quadratic equation $$n^2+n(-1-k^2)+(1-k-k^2)=0$$ Since these are all integers, the discriminant must be a perfect square, that is $$(k^2+1)^2-4(1-k-k^2)=(k^2+3)^2+4k-12$$ It is trivial to conclude that this cannot be a square greater than $(k^2+3)^2$ so we can only have $k=3$. Which gives us $n=11,p=37$.

Therefore, the only solutions are $(n,p)\in \{(1,2),(11,37)\}$.

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