The solutions of the equation $|p^3-q^2|=2^n$

Question

Let's consider the following equation $$|p^3-q^2|=2^n$$ where $p$, $q$ are distinct primes and $n$ a non-negative integer. The equation admits some trivial solutions: $$|2^3-3^2|=2^0$$ $$|3^3-5^2|=2^1$$ $$|5^3-11^2|=2^2$$ $$|17^3-71^2|=2^7$$ Clearly, not all values of $n$ admit a solution (for example, if $3|n$ the equation has no solution).

Are there other solutions besides those mentioned above?

Many thanks.

Answer 1

Thanks to pisco's comment, the answer is now complete.

First we can think about it as the families of elliptic curves $y^2=x^3\pm 2^n$, and instead of looking for prime coordinate points $(p,q)$, we can use the isomorphism $(x,y)\mapsto (4x,8y)$ and then divide out by $2^6$ so that we are instead looking at the finitely many curves where $0 \le n < 6$ but now focus on the points with $(\frac{p}{4^m}, \frac{q}{8^m})$ for some $m$.

In particular we have 8 curves of the form $y^2=x^3+a$ to look at, specifically for $a\in\{\pm 2^1, \pm 2^2, \pm 2^4, \pm 2^5\}$ since we already know $3 \not | n$. Then computing the rank of these curves with Sagemath we find they are all rank 0 except for $a \in \{2, -2, -4\}$. This immediately means the other 5 curves have only integer points and finitely many of them, so we can exhaust them with Nagell-Lutz. The integer points with distinct prime number coordinates on all these curves is: $(2,3), (3,5), (5,11)$

At this point we can use Sagemath to get the generators $(-1,1), (3,5), (2,2)$ for the curves with $a \in \{2, -2, -4\}$ respectively. Doing a brute force search, that gives the extra point $(\frac{17}{2^2}, \frac{71}{2^3})$ on $y^2=x^3+2$.

From here, using pisco's comment about S-integral points we have the full list.

Answer 2

We consider the elliptic curves $y^2 = x^3\pm2^n$ with parameter $n$ where $x$ and $y$ are prime number .
Searching the integer points $(x,y)$ with $\boldsymbol{n \lt 100}$ and $\boldsymbol{x \lt 10^{10}}$, we found the integer points $(x,y,n)=(3, 5, 1)$ and $(5, 11, 2)$ for $y^2 = x^3-2^n.$
Similarly, we found the integer point $(x,y,n)= (17, 71, 7))$ for $y^2 = x^3+2^n.$
There seems to be no other solution.