No nontrivial Syracuse cycle of length $4$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. Its only positive point of period dividing $4$ is $1$:

Equivalently, the Collatz map has no nontrivial cycle whose accelerated form has length $4$.

Mathematical role. With the corresponding statements for periods $1$, $2$ and $3$, this closes every accelerated cycle length up to $4$ unconditionally. Writing $x_0=m,\dots,x_3$ for the orbit and multiplying the four step relations $2^{v_i}x_{i+1}=3x_i+1$ gives

where $P=x_0x_1x_2x_3$, $K=\sum v_i$, and $e_j$ is the $j$-th elementary symmetric function of the orbit. The right-hand side is positive, so $2^{K}>81$, and the next power of two is $128$ — leaving the generous margin $2^{K}-81 \ge 47$. That margin is incompatible with all four orbit points being at least $3$, so the cycle meets $\{1\}$, and $1$ is fixed.

Why this is where the elementary method stops. The slack driving these proofs is the gap between $2^{K}$ and $3^{a}$. It is wide at $a=4$ — $128$ against $81$, a ratio of $1.58$ — which is why this case is in fact easier than $a=3$. At $a=5$ it collapses: $2^{8}=256$ against $3^{5}=243$, a ratio of $1.05$. That is not an accident. The admissible ratio satisfies $K/a > \log_2 3 = 1.5849\ldots$, and $8/5 = 1.6$ is a continued-fraction convergent of $\log_2 3$, so the power of two can crowd $3^{a}$ arbitrarily closely along the convergents. Excluding those periods is exactly a question about how well $\log_2 3$ is approximated by rationals, and no amount of elementary product manipulation replaces it.

Formalization note. No oddness hypothesis is needed: a point of period dividing $4$ under $T$ is automatically odd. The statement covers periods $1$, $2$ and $4$ simultaneously, since each divides $4$.