$23005\cdot (2^n-1)\cdot 2^n +1=p^2$ [closed]

Question

Consider:

$23005\cdot (2^n-1)\cdot 2^n +1=p^2$ , where n is an natural number and p a prime.

I conjecture that $p=9631$ is the only prime satisfying the above equation for $n=6$.

The curious fact is that $9631$ is the inverse of $37^2$.

Answer 1

Multiplying the equation by $2^n$ and denoting $X:=2^n$ and $Y:=p2^{\lfloor n/2\rfloor}$, we obtain two elliptic curves (depending on the parity of $n$): $$(23005(X-1)X+1)X=Y^2,$$ $$(23005(X-1)X+1)X=2Y^2.$$ Integral points on these curves can be easily computed with existing software (e.g., SageMath).

The former curve has integral points $(X,Y)\in\{ (0, 0),\ (1, \pm 1),\ (64, \pm 77048) \}$, while the latter has the only integral point $(X,Y)=(0, 0)$. From these integral points we conclude that the original equation has the only solution, which is $(n,p)=(6,9631)$.