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Let $ C : F(x,y,z)=0$ be a projective genus $1$ curve over $\mathbb{Q}$ with no restriction on the degree.

Write a point $P = (X , Y , Z)$ with the smallest coprime integers $X,Y,Z$.

Is it true that for every fixed $ a > 0$

$$ \log \max(|X|,|Y|,|Z|)- \log \min(|X|,|Y|,|Z|) > a $$ finitely often?

I believe it is true for homogenized Weierstrass model, false for genus $0$.

Limited experiments with cubics and quartics suggest it might be true.

Solution for degrees $3,4$ would be of interest too.

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I don't think so. Suppose that $T=(0,0)$ is a rational point on your curve $C$, and suppose that the rational points on $C$ lie dense around $T$ in the real topology. (It is easy to find such a $C$.) Your assertion would preclude $x=X/Z$ and $y=Y/Z$ from getting arbitrarily close to $(0,0)$, but this is exactly what happens for $x(P)$ and $y(P)$ if $P$ tends to $T$.

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  • $\begingroup$ Thanks. Would you please suggest explicit constructions of counterexamples? $\endgroup$
    – joro
    Commented Dec 12, 2013 at 14:17
  • $\begingroup$ With pleasure. All you need is an elliptic curve $C$ of the form $y^2=x^3+ax^2+bx$ (this visibly contains the point $T=(0,0)$) such that the rank of $C(\mathbb{Q})$ is $>0$ (which causes the rational points to lie dense in every connected component of $C(\mathbb{R})$ that contains rational points at all). $\endgroup$
    – Milton
    Commented Dec 12, 2013 at 14:20
  • $\begingroup$ BTW: the rank $>0$ condition can be achieved by choosing $a$ and $b$ to be integers such that $C$ contains a rational point $P_0$ with non-integral coordinates. By Nagell-Lutz (or actually by the theory of the formal group), $P_0$ is of infinite order, thus giving $C(\mathbb{Q})$ rank $>0$. $\endgroup$
    – Milton
    Commented Dec 12, 2013 at 14:22
  • $\begingroup$ Thanks. And take any generator of $C$? $\endgroup$
    – joro
    Commented Dec 12, 2013 at 14:25
  • $\begingroup$ I am not saying it is easy to write down a sequence of rational points $P_i$ that converges to $T$. (For this, it would probably be helpful to have an explicit isomorphism of Lie groups between $E(\mathbb{R})$ and $\mathbb{S}^1$ times a discrete group of order $1$ or $2$.) But what I am saying is: if $P_0$ has infinite order, then either $\{nP_0\}$ or $\{nP_0+T\}$ has a subsequence converging to $T$, and if $nP_0=(x_n:y_n:z_n)$ gets arbitrarily close to $T$, then $x_n$ and $y_n$ must get arbitrarily small compared to $z_n$, contradicting your displayed inequality. $\endgroup$
    – Milton
    Commented Dec 12, 2013 at 15:33

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