Small solutions of $x^2-a^3 y^2=\pm 1$

Question

We are interested in small integer solutions to the Pell equation: $$x^2-a^3 y^2=\pm 1 \qquad (1)$$

Where in $\pm 1$ you can chose either sign. $(x^2,a^3 y^2)$ are consecutive powerful numbers.

$abc$ implies that in the solutions of (1) $x$ can't be bounded by polynomial in $a$ for infinitely many $a$.

Q1 Can you unconditionally show that $x$ can't be bounded by $O(a^D)$ for all $D$ and infinitely many $a$?

For the simpler Pell equating $x^2- a y^2=1$ we can bound $x$ if $a=u^2-1$ and $x=u,y=1$.

One possible approach is continued fractions. Let $\frac{X_n}{Y_n}$ be the $n$-th convergent of the continued fraction of $\sqrt{a^3}$.

Let $N$ be the smallest integer such that $x=X_N,y=Y_N$ is solution to (1).

Q2 Must $N$ be unbounded for all $a$?

Computing $X_m,Y_m$ for $m \le 50$ is related to the integer sequence A135735 Sequence arising in a search for three consecutive powerful numbers.

Answer 1

It is better to ask one question per post. Here is the answer to Q1.

Assume that $(x,y)$ is a positive integer solution of $(1)$. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2-1$, where $u>1$ is an integer. Then the positive integer solutions of $X^2-aY^2=1$ are the pairs $(X_n,Y_n)$ satisfying $$X_n+Y_n\sqrt{a}=(u+\sqrt{a})^n,\qquad n\in\mathbb{Z}_{\geq 1}.$$ Hence $(x,ay)=(X_n,Y_n)$ for some positive integer $n$, which means that $$ay=Y_n=\sum_{\substack{0\leq k\leq n\\\text{$k$ odd}}}\binom{n}{k}u^{n-k}a^{(k-1)/2}.$$ On the right-hand side each term with $k>1$ is divisible by $a$, hence the term with $k=1$ is also divisible by $a$. That is, $a\mid nu^{n-1}$. So $a\mid n$, and then $$x=X_n>\frac{X_n+Y_n\sqrt{a}}{2}=\frac{(u+\sqrt{a})^n}{2}>\frac{(2\sqrt{a})^a}{2}>a^{a/2}.$$ So $x$ is not $a^{O(1)}$ when $a=u^2-1$.