Is $2^{n+1} \leq a + b \leq 4^n$ when $4^n - 1 = ab \quad \text{; with } n,a,b \in \mathbb{N}$? $4^n - 1 = ab \quad \text{; with } n,a,b \in \mathbb{N} \text{ such that } a, b \mid 4^n - 1$
Is this inequation always true: $2^{n+1} \leq a + b \leq 4^n$ ?
I already know that:
$4^n - 1 = \left(2^n\right)^2 - 1 = \left(2^n - 1\right) \left(2^n + 1\right)$
If it would be:
$a = \left(2^n - 1\right)$ and $b = \left(2^n + 1\right)$
then $a + b = 2^{n+1}$.
If it would be:
$a = 1$ and $b = 4^n - 1$
then $a + b = 4^n$.
But what about the other cases?
 A: Trying to do a proof by checking certain cases, as you're trying, is not sufficient as there's an infinite # of them. Instead, you need to use a more general technique. First, though, I assume by $\mathbb{N}$ you mean integers which are positive, i.e., it doesn't include $0$, since otherwise your inequalities are not always necessarily true (e.g., you could get $4^{0} - 1 = 0 = ab$, so $a = b = 0$, but $2^{0 + 1} = 2 \le 0 + 0 = 0$ is not true).
As you stated, you're trying check if it's always true that
$$2^{n+1} \leq a + b \leq 4^n \tag{1}\label{eq1A}$$
given that
$$4^n - 1 = ab \implies 4^n = ab + 1 \tag{2}\label{eq2A}$$
Note $4^n - 1$ is odd, so $a$ and $b$ are odd integers. Also, due to the symmetry in $a$ and $b$ in the problem, assume WLOG that $a \ge b$. First, checking the right side of the inequality, you want to see if
$$a + b \le 4^n = ab + 1 \tag{3}\label{eq3A}$$
If $b = 1$, then \eqref{eq3A} becomes $a + 1 \le a + 1$, which is obviously true. Next, if $b \ge 3$, you then have
$$ab + 1 \ge 3a + 1 = 2a + a + 1 \ge 2a + b + 1 \gt a + b \tag{4}\label{eq4A}$$
This shows \eqref{eq3A} also holds in that case too. Update: As bjorn's comment below states, it's easier to see this since $a + b \le ab + 1$ is equivalent to $(a - 1)(b - 1) \ge 0$.
Next, note all odd perfect squares are congruent to $1$ modulo $4$, but since $4^n - 1 \equiv 3 \pmod 4$, this means you have that $a \neq b$. Since have assumed $a \ge b$, this means, using \eqref{eq2A}, you have
$$\begin{equation}\begin{aligned}
a - b & \ge 2 \\
(a - b)^2 & \ge 4 \\
a^2 - 2ab + b^2 & \ge 4 \\
a^2 + 2ab + b^2 & \ge 4ab + 4 \\
(a + b)^2 & \ge 4(ab + 1) \\
(a + b)^2 & \ge 4(4^n) \\
(a + b)^2 & \ge 2^{2(n+1)} \\
a + b & \ge 2^{n+1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows the left side of the inequality in \eqref{eq1A} is also always true.
