How to solve the equation $$\left \lceil {\left ( 3 \over 2 \right )}^K \right \rceil = \left \lfloor {3^K - 1} \over {2^K - 1} \right \rfloor$$ for $K \in \mathbb N$.
Are there any other solutions except $K=1$?
My attempt:
LHS: I used that ${2^K}\not{|}{3^K}$ {does not divide) to write
$$\left \lceil {\left ( {3} \over {2} \right )}^K \right \rceil =
\left \lfloor {\left ( {3} \over {2} \right )}^K \right \rfloor + 1
$$
RHS: $$\left \lfloor {\left ( {3^K - 1} \over {2^K - 1} \right )} \right \rfloor = \left \lfloor {\left ( {3} \over {2} \right )}^K + {{{\left ( {3} \over {2} \right )}^K - 1 } \over {2^K - 1}} \right \rfloor = \left \lfloor {\left ( {3} \over {2} \right )}^K \right \rfloor + \left \lfloor \left \{ {\left ( {3} \over {2} \right )}^K \right \} + {1 \over {2^K}} {\left ( {{3^K - 2^K } \over {2^K - 1}} \right )} \right \rfloor $$ where {...} is the decimal part of the number. The second term is decreasing function from $\frac{1}{2}$ to 0 (in limit $K\to\infty$), so $\left \{ {1 \over {2^K}} {\left ( {{3^K - 2^K } \over {2^K - 1}} \right )} \right \} = {1 \over {2^K}} {\left ( {{3^K - 2^K } \over {2^K - 1}} \right )}$ $$\\$$
Together: $$1 = \left \lfloor \left \{ {\left ( {3} \over {2} \right )}^K \right \} + {1 \over {2^K}} {\left ( {{3^K - 2^K } \over {2^K - 1}} \right )} \right \rfloor$$ Or as an inequality: $$1 \le \left \{ {\left ( {3} \over {2} \right )}^K \right \} + {1 \over {2^K}} {\left ( {{3^K - 2^K } \over {2^K - 1}} \right )} $$
The term $\left \{ {\left ( {3} \over {2} \right )}^K \right \}$ has form ${r} \over {2^K} $, where $r \in \left \{ 1, 3, ..., 2^K - 1 \right \}$ (only odd numbers). $$\\$$ And here I am stuck.