No nontrivial Syracuse cycle of length $2$

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

Equivalently, the Collatz map has no nontrivial cycle whose accelerated form has length $2$.

Mathematical role. In the accelerated dynamics a Collatz cycle of length $a$ is a solution of $T^{a}(m)=m$, so together with the corresponding statement for $a=1$ this closes the two shortest cases of the cycle problem unconditionally. Writing $x = T(m)$ and multiplying the two step relations $2^{v_1}x = 3m+1$ and $2^{v_2}m = 3x+1$ gives

The right-hand side is positive, so $2^{v_1+v_2} > 9$; being a power of two it is therefore at least $16$, and the equation forces $7mx \le 3m+3x+1$. Since $mx \ge m+x-1$ for positive $m,x$, this collapses to $m+x \le 2$, leaving only $m=x=1$.

The point of interest is that no Diophantine input is needed. For a cycle of length $a$ the two 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 $a$ amounts to showing that this window contains no admissible integer $K$ — a question about rational approximations to $\log_2 3$. For $a \le 2$ the gap between $2^{K}$ and $3^{a}$ is necessarily coarse enough that plain integrality closes it, which is why these cases fall to elementary arithmetic while larger $a$ does not.

Formalization note. No oddness hypothesis is required; a point of period $2$ under $T$ is automatically odd. The statement is period dividing $2$: it also covers the fixed-point case, since $T(m)=m$ implies $T^2(m)=m$.