Prove that $(3/2)^n\bmod 1\equiv (2^n-1)/2^n $ has infinitely many solutions as $n \to\infty$

Question

The first part of the question is false and was proved by the answer vefore. The second part of the question asks to prove that $(3/2)^n \bmod 1\equiv (2^n-1)/2^n$ also has infinite solutions as $n \to\infty$.

Answer 1

This is not true. A result of Beukers from 1981 (combined with a tiny amount of computation) implies that $$ \| (3/2)^k \| > 2^{-0.9k}, $$ for all $k \geq 5$. Here $\|x\|$ denotes the distance from a real number $x$ to the nearest integer.

Answer 2

You can prove by induction that if $n$ is even and is exactly divisible by $2^k$ then

$$3^n \equiv 1 + 2^{k+2} \bmod 2^{k+3}$$

Whereas if $n$ is odd then

$$3^n \equiv 3 \bmod 2^3$$

If $n \ge 6$, then $n \ge \log_2(n) + 3 \ge k + 3 \ge 3$ and hence the congruence can never hold; one doesn’t need anything as sophisticated as Beuker’s result.