No nontrivial Syracuse cycle of length $5$

Let $T(n) = (3n+1)/2^{\,v_2(3n+1)}$ be the Syracuse map. Its only positive point of period dividing $5$ is $1$:

Equivalently, the Collatz map has no nontrivial cycle whose accelerated form has length $5$.

Mathematical role. With the cases $a \le 4$ this closes every accelerated cycle length up to $5$ unconditionally. Multiplying the five step relations $2^{v_i}x_{i+1}=3x_i+1$ around the orbit and writing $P=\prod x_i$, $K=\sum v_i$ gives

with $e_j$ the elementary symmetric functions of the orbit. Positivity of the right side forces $2^{K}>243$, hence $2^{K}\ge 256$ and a margin of $13$.

This case is markedly tighter than its predecessors. The margin $2^{K}-3^{a}$ is $47$ at $a=4$ but only $13$ here, because $8/5=1.6$ is a continued-fraction convergent of $\log_2 3 = 1.5849\ldots$, so $2^{8}=256$ sits unusually close above $3^{5}=243$. The consequence is not that the elementary method fails, but that it needs a larger threshold: the comparison $81e_4 + 27e_3 + 9e_2 + 3e_1 + 1 < 13P$ holds once every orbit point is at least $33$, whereas $3$ sufficed at $a=4$. The residual check is correspondingly larger — the sixteen odd values below $33$ — and none of them is $5$-periodic.

On how far this goes. Each individual period is decidable this way: for a given $a$, the margin determines a finite threshold $B_a$, and one checks the odd values below it. What does not follow is the general statement, which needs all $a$ at once, and the thresholds grow with the quality of the rational approximations to $\log_2 3$. Excluding all periods uniformly remains a question about linear forms in logarithms.

Formalization note. No oddness hypothesis is needed: a point of period dividing $5$ under $T$ is automatically odd. The statement covers period $1$ as well, since $1$ divides $5$.