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

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


(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.

  • $\begingroup$ Sadly $\mathbb{Q}[\sqrt{-5}]$ is not a UFD. $\endgroup$ – Captain Darling Apr 23 '13 at 20:50
  • 2
    $\begingroup$ @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

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 $(n,p)\in \{(1,2),(11,37)\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.