The only positive fixed point of the Syracuse map is $1$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map, sending $n$ to the odd part of $3n+1$. Its only positive fixed point is $1$:

Mathematical role. In the accelerated dynamics a Collatz cycle of length $a$ is exactly a point with $T^{a}(m)=m$, so this is the case $a=1$ of cycle exclusion, settled unconditionally. Writing $2^{v}$ for the power of two dividing $3m+1$, the fixed-point equation says $2^{v}m = 3m+1$, that is

which in the natural numbers forces $m=1$ and $2^{v}=4$ — the familiar $1 \mapsto 4 \mapsto 2 \mapsto 1$ loop of the classical map, collapsed to a single accelerated step.

This is the one case of the cycle problem that needs no information about how well $\log_2 3$ can be approximated by rationals. For $a \ge 2$ the analogous elimination requires the window $\log_2 3 < K/a \le \log_2(3+1/m)$ to contain no admissible integer $K$, which is a genuine Diophantine question; for $a=1$ the window argument degenerates and bare divisibility suffices.

Formalization note. No oddness hypothesis is needed: a fixed point of $T$ is automatically odd, since $T$ always returns an odd part. The valuation is written as Nat.factorization at the prime $2$.