Three Collatz steps decrease every $n \equiv 1 \pmod 4$ with $n > 1$

Let $C$ denote the Collatz step map. If $n > 1$ and $n \equiv 1 \pmod 4$, then exactly three steps of the map bring the orbit below its starting point:

Write $n = 4k+1$ with $k \ge 1$. Since $n$ is odd, the first step gives $3n+1 = 12k+4$, which is even; the second step gives $6k+2$, again even; and the third gives

The conclusion follows from $3k+1 < 4k+1 = n$, which holds precisely because $k \ge 1$, i.e. because $n > 1$. The value $n = 1$ must be excluded: it satisfies $1 \equiv 1 \pmod 4$ and lies on the cycle $1 \to 4 \to 2 \to 1$, so $C^{3}(1) = 1$.

Together with the even case, this lemma disposes of three of the four residue classes modulo $4$ in the study of Collatz descent, leaving only $n \equiv 3 \pmod 4$ open.