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Let $E$ be an elliptic curve over rational field. Let $P=(a/d^2,b/d^3)\in E(\mathbb{Q})$ and $$G=\{P,2P,3P,4P,\cdots\}.$$ Is there an integral point $Q\in G?$

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3 Answers 3

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I presume that (1) $E$ is meant to be in minimal form with integer coefficients, else you can change coordinates to put $P$ and $nP$ anywhere, and that (2) $P$ is not itself integral (i.e. $d>1$), else $Q=P$ works.

Then the answer is No, because $d^2$ and $d^3$ divide the denominators of $x(nP)$ and $y(nP)$ for all $n$.

This is a known property of "elliptic divisibility sequences". One way to think about it is that $p|d$ means that $P$ is $p$-adically near the identity point $\infty$ on $E$ (and $p^e|d$ for $e>1$ makes $P$ even nearer), and then any multiple of $P$ is at least as close.

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  • $\begingroup$ Noam, regarding your point (1), I think it's enough that the equation for $E$ have integer coefficients. It's not necessary that it be minimal in order to deduce that $d(P)\mid d(nP)$. $\endgroup$ Sep 30, 2014 at 21:17
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You probably meant to specify that the coordinates of $P$ are in lowest terms, and that $d\ge2$. Then, as Noam indicates, the answer is no. And it doesn't matter if the equation is minimal, as long as it has integer coefficients and as long as you start with a point $P$ having non-integer coordinates.

A good way to study this question is via the formal group, see for example Chapters 4 and 7 of my book The Arithmetic of Elliptic Curves. Assuming that you are on a minimal Weierstrass equation, one can prove something even stronger, namely if $p\mid d$ and if we write $nP=(a_n/d_n^2,b_n/d_n^3)$ in lowest terms, then the power of $p$ dividing $d_n$ is exactly $$ \operatorname{ord}_p(d_n) = \operatorname{ord}(d) + \operatorname{ord}(n). $$ This shows that $d\mid d_n$ for every $n$, as Noam indicated, but is considerably more precise.

Lots more is known about the sequence $(d_n)$, which is called an elliptic divisibility sequence. For example, there is an $N$ so that for every index $n>N$ the term $d_n$ in the sequence is divisible by a prime that doesn't divide any of the earlier $d_i$ with $i<n$. On the other hand, it is not known whether there are infinitely many primes $q$ such that if $d_n$ is the first term in the sequence divisible by $q$, then $q^2\mid d_n$. (These are the elliptic analogue of Wieferich primes.)

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Not necessarily. In fact, if the answer was yes, there would be an easy (yet time consuming) algorithm for computing Mordell-Weil ranks, which we don't have.

The lowest conductor example with positive rank and no integral points is this elliptic curve.

sage: E = EllipticCurve('189b3')
sage: E.rank()
1
sage: E.gens()
[(-143/4 : -3/8 : 1)]
sage: E.integral_points()
[]

I looked it up by typing this into sage:

sage: for E in cremona_curves(1..1000):
    if E.rank() > 0:
        if len(E.integral_points())<3:
            print E.cremona_label(), E.rank(), E.integral_points()

This functionality should be helpful in answering a lot of other questions of this type you might have about elliptic curves. Some can even by answered simply by using the search feature on LMFDB, but this isn't one of them. You can read more about Cremona's database (now up to conductor at least 300,000) in this report.

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  • $\begingroup$ Thanks, but I mean is it possible? Indeed I'm looking for an example, which there is such integral point! I can't find such example!!! $\endgroup$ Sep 30, 2014 at 15:45

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