Integer solutions of $n^3 = p^2 - p - 1$ Find all integer solutions of the equation, $n^3 = p^2 - p - 1$,  where p is prime.
 A: 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.
A: (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.
A: 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 $(n,p)\in \{(1,2),(11,37)\}$.
