For which positive integers $k$ dividing $4p^3$, does the equation $x^3+px=k^2$ has an integer solution in $x$? Let $p$ be an odd prime. For which positive integers $k$ dividing $4p^3$, does the equation $x^3+px=k^2$ has an integer solution in $x$ ?
Since for integer $x$, $px+x^3$ is even, so $k$ is even, so $4|x^3+px$. Now if $x$ is even, then this implies $4|x$. 
I am unable to deduce anything else. Please help. 
 A: The question as stated is really just 12 different Diophantine equations:
\begin{align*}
&x^3+px=1 & &x^3+px=4 & &x^3+px=16 \\
&x^3+px=p^2 & &x^3+px=4p^2 & &x^3+px=16p^2 \\
&x^3+px=p^4 & &x^3+px=4p^4 & &x^3+px=16p^4 \\
&x^3+px=p^6 & &x^3+px=4p^6 & &x^3+px=16p^6
\end{align*}
As already noted, the LHS of each equation is even, so the 4 equations on the left have no solutions. Since $p$ is an odd prime, there can be no solutions to any of the remaining 8 Diophantine equations involving non-positive values for $x$, since the LHS would be non-positive while the RHS is positive.
Thus, the equation $x^3+px=4$ has only one solution $(x,p)=(1,3)$, and the equation $x^3+px=16$ has no solutions.
For the remaining 6 Diophantine equations, it is clear that any solution must have $x$ divisible by $p$. It is obvious that no solution to $x^3+px=4p^2$ or $x^3+px=16p^2$ can have $x\geq 2p$, and that there are only finitely many primes $p$ to check; we then see that $x^3+px=4p^2$ has only one solution $(x,p)=(3,3)$ and that $x^3+px=16p^2$ has no solutions.
Now consider the equation $x^3+px=4p^4$. Any solution must not only have that $x$ is divisible by $p$, but in fact that $x$ is divisible by $p^3$. Thus, if there is a solution, defining $y=p^3x$ and substituting into our equation yields that $p^9y^3+p^4y=4p^4$ must also have a solution. This is clearly absurd.
The exact same argument works for the equation $x^3+px=16p^4$, and the final 2 equations are handled with similar arguments (instead of divisibility by $p^3$ you get divisibility by $p^5$).
Therefore, the only solutions to any of the 12 Diophantine equations in consideration are $(x,p)=(1,3)$ for $x^3+px=4$ and $(x,p)=(3,3)$ for $x^3+px=4p^2$.
