I encountered this question and have been stuck for couple of days.
Solve in integers
$$1)\quad x^2=y^6+24y^3+192$$ $$2) \quad x^3=y^6+24y^3+192$$
(Two different diophantine equation.)
Seems one solution for 1. is $y=-1$ and $x=13$.
I have tried modular arithmetic and but nothing seems to work so far. I've only been able to reduce it into an equivalent Diophantine Equation i.e.,
Any help will be appreciated
Best Answer
For the first: rewrite as
$$x^2 = (y^3 + 12)^2 + 48.$$
The squares differing by 48 are
\begin{align*} (1,49) \\ (16,64) \\ (121 ,169) \\ \end{align*}
which you can get by solving $(z + 6)^2 = z^2 + 48$, $(z + 4)^2 = z^2 + 48$ and $(z + 2)^2 = z^2 + 48$ respectively.
Choosing $x$ to get the right hand squares should pose no trouble. $(y^3 + 12)^2 = 1$ has no integer solution, $(y^3 + 12)^2 = 16$ has $y = -2$ as a solution and $(y^3 + 12)^2 = 121$ has $y = -1$ as a solution.
So the solution set is \begin{align*} (x,y) &= (8, -2) &\lor\\ (x,y) &= (-8, -2) &\lor\\ (x,y) &= (13, -1) &\lor \\ (x,y) &= (-13, -1) \end{align*}