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I hope this question is ok, to post it here, otherwise I will move it to MSE:

Let $y^2 = x^3+ax+b$ be an elliptic curve $E$ where $a,b \in \mathbb{Z}$. Let $M_x = \max_{x}|x|$ and $M_y = \max_{y}|y|$ where the maximum runs through the integral points $P=(x,y)$ of $E$. If there are no integral points set $M_x = M_y = 0$. Let $N(E)$ be the number of integral points. I have done some small experiments in Sagemath for $a=-n^2,b=n^2, 1 \le n \le 100$ and there seems to hold this inequality:

$$ N(E) < 2 \log( (M_x+1)(M_y+1)+1)$$

  1. Question:

Is there any reason or heuristic this could be true or are there elliptic curves where this is not true?

Here is a picture: On the x-axis are the number of integral points, on the y-axis is the upper bound above:

upper_bound_number_of_integral_points_on_elliptic_curve

More background why I am interested in this:

Let $R = \operatorname{rad}(\gcd(a,b))$

Then for each prime $p | \gcd(a,b)$ we have

$$y^2 \equiv x^3 \mod(p)$$

The number of such solutions $N_p$ is $\ge p$ as $(x,y) = (t^2,t^3)$ for each $t \in \mathbb{F}_p$ are such solutions. My conjecture, which I think can be proven since $\mathbb{F}^*_p$ is cyclic, is that $N_p = p$.

From this it follows that for each integral point $(x,y)$ on $E$, we can find $t,k,l \in \mathbb{Z}$ such that (if $R=1$ set $t=0$):

$$x = t^2+k R$$

$$y=t^3+lR$$

In fact, using the Chinese Remainder Theorem, given an integral point $(x,y)$ on $E$, we can compute:

$$\sqrt{x} \equiv t_i \mod(p_i)$$ $$(y)^{\frac{1}{3}} \equiv t_i \mod(p_i)$$

for $i=1,\cdots,r$ ,where $R = p_1\cdots p_r$. And we can find $0 \le t \le R-1$.

From this we get upper bounds for $|x|,|y|$:

$$|x| \le R(R+|k|)$$

and using the elliptic equation we get:

$$|y| \le \sqrt{R^3(R+|k|)^3 + |a|R(R+|k|)+|b|}$$

Now comes the cheating:

Let $M_k := \max_{k} |k|$ and set

$$B_x := R(R+M_k)$$

$$B_y := \sqrt{R^3(R+M_k)^3 + |a|R(R+M_k)+|b|}$$

With the notation of above it is $M_x \le B_x$ and $M_y \le B_y$.

  1. Question: Is there any way that we can upper bound $M_k$, if this is not asked too much?

Thanks for your help!

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An LMFDB search for curves with many integer points turns up the curve 20888a1: $y^2 = x^3 - 52 x + 100$ which has $52$ integral points, a bit more than your conjectured bound of about $47.052$ using $(M_x, M_y) = (12214, 1349854)$.

It does seem to be true that curves with many integer points tend to have a few large ones, but I don't know of any quantitative statement in this direction.

There are known upper bounds on the size of an integral point on an elliptic curve, but they are usually quite large (though still useful for provably listing all integer points).

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