We know that $n^2$ must be close to $N^3$ and $(n+999)^2$ close to $(N+1)^3$. By setting $m^6=N^3=n^2$ we get the equation

$$3m^4-1998m^3+3m^2-998000=0$$

which gives the feasible root $m\approx666$.

Now the solution is close to $N=666^2$. The next perfect square is $n^2:=(\lfloor\sqrt{N^3}\rfloor+1)^2$ and we must fulfill

$$(n+999)^2<(N+1)^3\le(n+1000)^2$$

this can be solved by trial and error around $N=666^2$.

We know that $n^2$ must be close to $N^3$ and $(n+999)^2$ close to $(N+1)^3$. By setting $m^6=N^3=n^2$ we get the equation

$$3m^4-1998m^3+3m^2-998000=0$$

which gives the feasible root $m\approx666$.

Now the solution is close to $N=666^2$. The next perfect square is $n^2:=(\lfloor\sqrt{N^3}\rfloor+1)^2$ and we must fulfill

$$(n+999)^2<(N+1)^3\le(n+1000)^2$$

This can be solved by trial and error around $N=666^2$.

We know that $n^2$ must be close to $N^3$ and $(n+999)^2$ close to $(N+1)^3$. By setting $m^6=N^3=n^2$ we get the equation

$$3m^4-1998m^3+3m^2-998000=0$$

which gives the feasible root $m\approx666$.

Now the solution is close to $N=666^2$ and must fulfill

$$(\lfloor\sqrt{N^3}\rfloor)^2\le N^3<(\lfloor\sqrt{N^3}\rfloor+1)^2,(\lfloor\sqrt{N^3}\rfloor+1000)^2<(N+1)^3\le(\lfloor\sqrt{N^3}\rfloor+1001)^2$$

We know that $n^2$ must be close to $N^3$ and $(n+999)^2$ close to $(N+1)^3$. By setting $m^6=N^3=n^2$ we get the equation

$$3m^4-1998m^3+3m^2-998000=0$$

which gives the feasible root $m\approx666$.

Now the solution is close to $N=666^2$. The next perfect square is $n^2:=(\lfloor\sqrt{N^3}\rfloor+1)^2$ and we must fulfill

$$(n+999)^2<(N+1)^3\le(n+1000)^2$$

this can be solved by trial and error around $N=666^2$.