Solving A Simple Exponential floor Equation

Question

So, there's this equation that I've been pondering upon:(to slove for n) $\left\lfloor N\alpha ^{n} \right\rfloor = \left\lfloor N\alpha ^{n+1}\right\rfloor+1$
I can't remove the floors for approximation purposes because that would distort the answer too much. But, I thought of using the definition of floors.That is if
$\left\lfloor k\right\rfloor = m$
Then:
$m\leq k < m+1$.
Thus, here we can say that we've
$\left\lfloor N \alpha ^{n+1}\right\rfloor+1 \leq N\alpha ^{n} < \left\lfloor N \alpha ^{n+1}\right\rfloor+2$
But that would only complicate the matters more.
Thus, I'm stuck down here. Any help appreciated.

Answer 1

Hint: you are on the right track with thinking at the definitions. Write it as:

$$N \alpha^{n+1} \lt \lfloor N \alpha^n \rfloor \le N \alpha^n \lt \lfloor N \alpha^n \rfloor + 1 = \lfloor N \alpha^{n+1} \rfloor + 2 \le N \alpha^{n+1} + 2$$

$$ \implies \quad N \alpha^{n+1} \lt N \alpha^n \lt N \alpha^{n+1} + 2 \quad \iff \quad \alpha \lt 1 \lt \alpha + \frac{2}{N \alpha^n} $$