I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$ I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$?
I tried to solve it as follows:
I state $a\in\mathbb{N}$, so that $\sqrt{x^2+2x+4}<a<\sqrt{4x^2+2x+1}$
Hence $x^2+2x+4<a^2<4x^2+2x+1$
$4x^2+2x+1-x^2-2x-4=3x^2-3=3*2^{4004}-3$.
Here is where I got stuck. Could you please explain to me how to solve this question?

With the lower limit,
$$\begin{equation}\begin{aligned}
& (x + 1)^2 \lt x^2 + 2x + 4 \lt (x + 2)^2 \\
& x + 1 \lt \sqrt{x^2 + 2x + 4} \lt x + 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This means, with $a \in \mathbb{N}$ as you stated,
$$a \ge x + 2 \tag{2}\label{eq2A}$$
With the upper limit,
$$\begin{equation}\begin{aligned}
& (2x)^2 \lt 4x^2 + 2x + 1 \lt (2x + 1)^2 \\
& 2x \lt \sqrt{4x^2 + 2x + 1} \lt 2x + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This thus means
$$a \le 2x \tag{4}\label{eq4A}$$
Combining \eqref{eq2A} and \eqref{eq4A} gives
$$x + 2 \le a \le 2x \tag{5}\label{eq5A}$$
Thus, as zwim's hints state, the number of integers in between the $2$ square root limits would be
$$2x - (x + 2) + 1 = x - 1 = 2^{2002} - 1 \tag{6}\label{eq6A}$$
