Solutions of $y^2=x^{2n+1}-3(4^{2n})$?

Question

I would like to prove that in the following Diophantione equation $$ y^2=x^3-48,\qquad (x,y\in\mathbb{Z}^+) $$ has a solution $(x,y)=(4,4)$ and $(x,y)=(28,148)$ and no other solutions.

Also for any $n\ge2$ the follwing Diophantine equation $$ y^2=x^{2n+1}-3(4^{2n}),\qquad (x,y\in\mathbb{Z}^+) $$ has only a solution $(x,y)=(4,4^n)$ and no other solutions.

Thanks.

Answer 1

$y^2=x^3-48$ is solved in Mordell, Diophantine Equations. The proof is too long for me to write out here. But here's some of what Mordell says, on page 247:

"...the equation $y^2=x^3-48$ ... has only the solutions $(x,y)=(4,\pm4),(28,\pm148)$ and is of particular interest. On replacing $x$ by $4x$ and $y$ by $8y+4$, it becomes $$y^2+y+1=x^3=(y-\rho)(y-\rho^2),$$ which has only the solutions $(x,y)=(1,0),(1,-1),(7,18)$. This follows since from the equation $$y-\rho=\pm\rho^{\alpha}(p+q\rho)^3,$$ we easily deduce an equation $$X^3-3XY^2+Y^3=1.$$ Ljunggren solves this completely in the manner indicated above, and some details are given in Chapter 23."

The Ljunggren reference is Einige Bemerkungen uber die Darstellung ganzer Zahlen durch binare kubische Formen mit positive Diskriminante, Acta Math 75 (1942) 1-21."