# Upper bounds for number of integral points on short Weierstraß elliptic curve?

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:

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?

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 $$(M_x, M_y) = (12214, 1349854)$$.