Exactly $1000$ perfect squares between two consecutive cubes

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 $N=666^2$.

(This is a redraft to be clearer. And also to fix @mathematician123's point about the possibility that $N+1$ is a perfect square.)

I'm going to do this for any large enough number $n$ of perfect squares. And then show that $1000$ is large enough.

Let $g(x) = x^{3/2}$. Then the number of perfect squares below $N^3$ is $\lfloor g(N) \rfloor$. What we want to show is that there exists an interval of length at least $3$, inside of which $n < g'(x) < n + 0.5$. In that interval we will find three consecutive integers $N$, $N+1$, and $N+2$. Of them, either there are $n$ perfect squares between $N^3$ and $(N+1)^3$, or else there must be $n$ perfect squares between $(N+1)^3$ and $(N+2)^3$.

Let's start. $g'(x) = 1.5 \sqrt{x}$. $g'(0) = 0$, $g'(x)$ is increasing, continuous, and goes to infinity. Therefore for any positive integer $n$ there is always a solution to $g'(x) = n$.

$g''(x) = \frac{0.75}{\sqrt{x}}$. This is positive, decreasing, and goes to $0$ in the limit. By inspection, $g'(21) < 7$ and $g''(21) < \frac{1}{6}$. If $7 \le n$ let $x_0$ be the solution to $g'(x_0) = n$. $x_0$ is in the interval $(21, \infty)$ and therefore $0 < g''(x) < \frac{1}{6}$ in the interval $(x, x+3)$. By the intermediate value theorem $g'(x)$ cannot change by more than $3 * \frac{1}{6} = 0.5$, so in that interval $n < g'(x) < n + 0.5$.

Let $N$ be the smallest integer bigger than $x_0$. Then $N$, $N+1$ and $N+2$ are all in the interval $(x_0, x_0+3]$. By the intermediate value theorem, $g(N)+n < g(N+1) < g(N)+n + 0.5$. Likewise $g(N+1)+n < g(N+2) < g(N+1) + n + 0.5$. Therefore the triple $g(N), g(N+1), g(N+2)$ has the form $g(N), g(N) + n + \epsilon_1, g(N) + 2n + \epsilon_2$ where $0 < \epsilon_1 < \epsilon_2 < 1$.

Now there can be at most one integer in the interval $[g(N), g(N) + \epsilon_2]$. If there are none or it occurs after $g(N) + \epsilon_1$, then $\lfloor g(N+1) \rfloor - \lfloor g(N) \rfloor = n$ and those perfect squares are strictly between the cubes. If it occurs before $g(N) + \epsilon_1$ then $\lfloor g(N+2) \rfloor - \lfloor g(N+1) \rfloor = n$ and those perfect squares are strictly between the cubes. If it occurs exactly at $g(N) + \epsilon1$, then $\lfloor g(N+1) \rfloor - \lfloor g(N) \rfloor = n+1$, but one of those perfect squares is at $(N+1)^3$, and so there are exactly $n$ perfect squares strictly between $N^3$ and $(N+1)^3$. In all cases, the result holds.

This shows that if $7 \le n$, there is always an $N$ with $n$ perfect squares between $N^3$ and $(N+1)^3$. Since $7 < 1000$, this solves the problem.

As @TonyK already pointed out in the comments, examples of $N$ can be found for $n = 2, 3, 4, 5, 6, 7$. Therefore this actually holds for $2 \le n$.

This argument can be generalized to looking for counts of perfect $k$ powers between perfect $r$th powers as long as $k < r$. Though, of course, with different lower bounds on $n$.

I can't resist giving this as an answer, but the credit goes to @mathematician123 who put me on the right trail by his comment. So I will make this a Community wiki answer.

Claim: for $m>0$ an integer, the number of squares between the successive cubes $m^6=(m^2)^3$ and $(m^2+1)^3$ equals $\left\lfloor \frac{3m}{2} \right\rfloor$.

Since the smallest square $>m^6$ is $(m^3+1)^2$, it suffices to show the inequalities $$ \left( m^3+\left\lfloor \frac{3m}{2} \right\rfloor\right)^2 < (m^2+1)^3 < \left( m^3+\left\lfloor \frac{3m}{2} \right\rfloor + 1 \right)^2. $$ We will prove these by proving the following stronger inequalities for $m$ large enough, dealing with the small cases separately: $$ \left( m^3+\frac{3m}{2}\right)^2 < (m^2+1)^3 < \left( m^3+ \frac{3m}{2} + \frac{1}{2} \right)^2. $$ So let us first prove $\left( m^3+\frac{3m}{2}\right)^2 < (m^2+1)^3$. Expanding the LHS gives $m^6+3m^4+\frac{9}{4}m^2$, which is indeed smaller than $(m^2+1)^3=m^6+3m^4+3m^2+1$ in virtue of $3>\frac{9}{4}$.

Now we prove $(m^2+1)^3 < \left( m^3+ \frac{3m}{2} + \frac{1}{2} \right)^2$. Expanding the RHS of this, we get $$ \left( m^3+ \frac{3m+1}{2} \right)^2 = m^6+m^3(3m+1)+\left(\frac{3m+1}{2}\right)^2 = m^6+m^3(3m+1)+\frac{9}{4}m^2+\frac{3m}{2}+\frac{1}{4}=m^6+3m^4+m^3+\frac{9}{4}m^2+\frac{3m}{2}+\frac{1}{4} $$ Comparing this to the LHS and cancelling terms, we see that the inequality to be proved is equivalent to $$ m^3+\frac{3m}{2} > \frac{3m^2}{4}+\frac{3}{4} $$ This is clearly satisfied for $m\geq 2$. We can check the case $m=1$ of the claim by hand. The claim is proven.

Therefore it follows that if for some $m$ we have $\left\lfloor \frac{3m}{2} \right\rfloor = 1000$, it follows that there are $1000$ squares between the successive cubes $(m^2)^3$ and $(m^2+1)^3$. Since $1000$ is $1$ mod $3$, we can indeed write it in this form, and indeed $m=667$ works.

Therefore there are exactly $1000$ squares between the successive cubes $(667^2)^3$ and $(667^2+1)^3$, or between $444889^3$ and $444890^3$.

Finally, we can verify all of this by using the command line utility bc:

$ bc sqrt((667^2)^3) 296740963 sqrt((667^2+1)^3-1) 296741963 

$N = 443598$ works.

$$(295450254)^2 \leq N^3 < (295450255)^2$$ $$(295451254)^2 < (N+1)^3 \leq (295451255)^2$$

Given $$N^3\lt n^2\lt (n+1)^2\lt\cdots\lt(n+999)^2\lt (N+1)^3$$ we have $$\begin{cases}N^3\lt n^2\\(n+999)^2\lt (N+1)^3\end{cases}\implies1998n+999^2\lt3n^{\frac43}-1998n^\frac33+3n^{\frac23}-998000$$ Making $t=n^{\frac13}$ we get $$3t^4-1998t^3+3t^2-998000\gt0$$ Equating to $0$ we have in Desmos two roots of distinct sign and the positive root is $t=665.99962$ so $n\ge295407790.343$ while in Wolfram asking about the inequality $$(x+999)^2\lt (x^{\frac23}+1)^3$$ give the value $n\ge295\space 408\space 000$. Thus $$\boxed{n\ge295\space407\space790\space\text{in Desmos}\\n\ge295\space408\space000\space\text{in Wolfram}}$$ Trying to calculate with small values (less than $1000$) we can verify that for a certain number of squares there are several solutions. And this is so for the obtained values which, obviously should have an upper or major terminal.Determine it is question of calculation with computer.

The number of squares between $a$ and $b$ is approximately $\sqrt{b} - \sqrt{a}$. We're given that $a = N^3$ and $b = (N+1)^3$ with 1000 squares between them, so:

$$\sqrt{(N + 1)^3} - \sqrt{N^3} = 1000$$ $$\sqrt{N^3 + 3N^2 + 3N + 1} - \sqrt{N^3} = 1000$$

Squaring gives:

$$N^3 + 3N^2 + 3N + 1 - 2\sqrt{N^6 + 3N^5 + 3N^4 + N^3} + N^3 = 1000000$$ $$2N^3 + 3N^2 + 3N - 999999 = 2\sqrt{N^6 + 3N^5 + 3N^4 + N^3}$$

Squaring again gives:

$$(2N^3 + 3N^2 + 3N - 999999)^2 = 4(N^6 + 3N^5 + 3N^4 + N^3)$$ $$4 N^6 + 12 N^5 + 21 N^4 - 3999978 N^3 - 5999985 N^2 - 5999994 N + 999998000001 = 4N^6 + 12N^5 + 12N^4 + 4N^3$$ $$9N^4 - 3999982 N^3 - 5999985 N^2 - 5999994 N + 999998000001 = 0$$

We can obtain an approximate solution by ignoring all but the first two terms of the polynomial.

$$9N^4 - 3999982 N^3 \approx 0$$ $$N \approx \frac{3999982}{9} = 444442.444444\dots$$

In fact, the two real solutions to the polynomial equation are $N \approx 62.4950603534606$ and $N \approx 444443.9444444913$. But there are only 12 square numbers between $62^3$ and $63^3$, so that's not the answer we're looking for. Let's focus instead on $N$'s close to $444444$.

Searching by computer, I found that there are in fact 888 different values of $N$ that satisfy the requirement of having exactly 1000 perfect squares between $N^3$ and $(N+1)^3$.

For example, $N = 444444$ yields all the squares between $296295852^2$ and $296296851^2$, inclusive.