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:

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

- 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 $M_y \le B_y$.

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

Thanks for your help!