No nontrivial Syracuse cycle of length $3$

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

Equivalently, the Collatz map has no nontrivial cycle whose accelerated form has length $3$.

Mathematical role. Together with the corresponding statements for periods $1$ and $2$, this closes the three shortest cases of the cycle problem unconditionally. Writing $x_0=m$, $x_1=T(m)$, $x_2=T^2(m)$ and multiplying the three step relations $2^{v_i}x_{i+1}=3x_i+1$ gives

where $K=v_0+v_1+v_2$. The right-hand side is positive, so $2^{K}>27$, and a power of two exceeding $27$ is at least $32$; hence the left-hand side is at least $5\,x_0x_1x_2$. That inequality cannot hold when all three orbit points are $\ge 7$, since then the cubic term dominates the quadratic and linear ones. So the cycle meets $\{1,3,5\}$, and of those only $1$ is periodic: $3 \mapsto 5 \mapsto 1$ and $5 \mapsto 1$ both fall into the trivial cycle.

The interest is again that no Diophantine input appears. For a cycle of length $a$ the general bounds confine the ratio of even to odd steps to $\log_2 3 < K/a \le \log_2(3+1/m)$, and eliminating a given $a$ means showing this window holds no admissible integer $K$ — a question about rational approximations to $\log_2 3$. For small $a$ the gap between $2^{K}$ and $3^{a}$ is necessarily coarse ($32$ against $27$ here), leaving enough slack for plain integrality. That slack shrinks quickly as $a$ grows — at $a=5$ the gap is only $256-243=13$ — so this elementary ladder reaches a few more cases at most and does not extend to all periods.

Formalization note. No oddness hypothesis is needed: a point of period dividing $3$ under $T$ is automatically odd, since $T$ returns an odd part. The statement covers periods $1$ and $3$ simultaneously, as period $1$ divides $3$.