The so-called "1-cycle" in the Collatz-problem was already disproved by Ray Steiner 1977. However, he used transcendental number theory to achieve that, and Lagarias commented, it is surprising that such heavy weapons are needed for such a tiny result.
I think I have an elementary proof now but usually in the Collatz-problem your proofs are faulty, so I better ask for a crosscheck.
As proposed in the MSE.meta I'll put the problem-description in this question and post my attempted proof in an own answer.
The notion of 1-cycles begin with the definition of increasing steps $a_{k+1} = {3a_k +1\over 2} $ if $a_k$ is odd and decreasing steps $a_{k+1} = {a_k\over 2} $ if $a_k$ is even.
If we consider the simple type of a cycle, which occurs if $N$ increasing steps are followed by $B$ decreasing steps, then let us call this with Steiner, Simons, de Weger and others the "1-cycle" of length $S=N+B$.
Notes: 1) here the $N$ contains the number of increasing steps and defines, that the term $3^N$ will become involved. It is my main varying parameter.
2) Depending on $N$, there is $S$ such that $2^S \gt 3^N \gt 2^{S-1}$ so in fact $S$ is a function of $N$, and can also be determined by $S(N)=\lceil N \log(3)/\log(2) \rceil$ . I used the symbol $S$ here to indicate it is the sum of the exponents of all powers of $2$ which are involved and also the shorter form $S$ instead $S(N)$ .
3) I introduce also the short-name for the difference $B = S-N$, thus also depending on $N$ . $B$ is simply the number of purely decreasing steps after the $N$ purely increasing steps.
With Steiner, we can state, that an initial number a which we wish to increase by N steps must have the form $a = 2^N \cdot k -1$ ; here we can use any odd integer $k \gt 0$ . The result of that transformation, let's call it b has then the form $b=3^N \cdot k -1 $ with the same parameter k.
If b is then transformed by B decreasing steps and reaches c then c has the form $c={3^N \cdot k\ – 1 \over 2^B}$
If we want a cycle then this requires $c=a$ and this is also the "1-cycle" where $a$ is the minimal element.
Well, thus identifying a with c this gives the "critical formula"
$$ \tag{1.0} 2^N\cdot k – 1 = {3^N \cdot k -1 \over 2^B}$$
Here we assume some $N$, get from this the dependend variable $B$ and from this try to find some or one admissible $k$.
Simply rearranging gives the more focused form which appears (using different letters) in J.Simon's paraphrase of R.Steiner's proof:
$$ \tag{1.1} (2^S – 3^N)\cdot k = 2^B – 1 $$
(remember $S=N+B$ and $2^S \gt 3^N$ is understood). Then the relevant statement is:
For a 1-cycle of any length $N$ to exist we must have a solution in positive integers $(N,S,B,k)$ in eq. (1.1) .
$\qquad \qquad$ (two other useful representations of that critical equation are
$$ \small (2^S – 3^N)\cdot k = (2^S – 2^N )/2^N \quad , 2^S \gt 3^N \quad \text{ or }\\
\small 2^B = {2^S\over 2^N}={3^N \cdot k -1 \over 2^N\cdot k-1}$$
Using results of transcendental number theory and rational approximation with linear forms of logarithms Steiner proved, that there is no $2^B$ satisfying that formula except if $N=1$ and from this $S=2$,$B=1$,$k=1$ and $ a=1 \Rightarrow b = 2 = {3a+1\over 2} \Rightarrow a=1={b \over 2} $ which is called the "trivial cycle".
Question: how can it be proven by elementary means, that the formula (1.1) has not other solutions?
Actually, this question is asking for help: to crosscheck my proof attempt, which I'll state in an answer below.
The Steiner-proof is not online, but John Simons has used it when he extended the Steiner-method for the 2-cycle. His article which contains a paraphrase is online at AMS . Note that I use different letters for the variables to be consistent with my own older discussions
[Added]: Empirically, the last shown form of the critical equation can never be satisfied for $N \gt 1, k \ge 1$ because the rhs falls empirically between the sharp bounds
$$ \small {3^N\over 2^N} \lt {3^N \cdot k -1 \over 2^N\cdot k-1} \lt \Big\lceil {3^N \over 2^N} \Big\rceil \qquad \text{ for }K=1 \ldots \infty$$
(and thus is never arriving the next integer above $(3/2)^N$) where the right inequality conforms with that conjectured bound in the Waring-problem, but which is not yet proven.
Best Answer
[update2]: It was found, that the given considerations below are not completely sufficient to disprove the existence of a "nontrivial" solution in eq (1.1) and thus the disprove of a nontrivial cycle, as asked for in the question. However some modular restrictions are formulated
(Thanks to Steven to point out the flawed argument)[/update2]
I begin with the formula (1.1) in my question $$ (2^S - 3^N)\cdot k = 2^B - 1$$ We check modular conditions on $k,2^B$ and $3^N$ for compatibility.
We begin assuming there might a solution be existent, where also $2^S = 2^N \cdot 2^B \gt 3^N$ and thus $S \gt N \cdot \log_2 3 \approx N \cdot 1.59 $ and thus $B \gt N \cdot 0.59 $
We look at k first:
Corol.: $k \lt 2^B-1$ for any nontrivial case $N \gt 1$.
Proof: It is obvious that $ k \le 2^B-1 $ so we check, whether $k = 2^B-1$ . This gives $ (2^S - 3^N)\cdot (2^B - 1) = 2^B - 1$ and then $(2^S - 3^N) = 1 $
But because consecutive powers of 2 and 3 cannot have the difference 1 except for $2^2$ and $3^1$ we must have
$ \qquad k \lt 2^B-1 $
Corol.: $k \gt 1 $ for any nontrivial cycle $N \gt 1$
Proof: Assume $k=1$. Then the equation can be rewritten as
$ \qquad 2^S - 3^N = 2^B - 1 $
$ \qquad 2^S - 2^B = 3^N - 1 $
$ \qquad 2^B \cdot (2^N - 1) = 3^N - 1 $
We look at N:
a. if N is even, then the parenthese on the lhs has the primefactor 3 but not the rhs, so N cannot be even
b. if N is odd, then the parenthese on the rhs has the primefactor 2 only to the first power, so $B=1$ and we're again arriving at the "trivial cycle" as only solution
Corol.: $ 3 \le k \le 2^B-3 $ : the possibility of an integer-solution for the main equation depends on modular restrictions on the residue $3^{-N} \pmod{2^B}$, and thus the possibility of a nontrivial cycle.
Proof:
First k cannot be even because $2^B-1$ is odd so $ 3 \le k \le 2^B-3 $.
Next we write the numbers in modular terms $ \pmod{2^B}$
$\qquad 2^S = 2^B \cdot 2^N$
$\qquad 3^N = 2^B \cdot n + r $ with necessarily $ n \le 2^N - 1$ and also $0 \lt r \lt 2^B$
$\qquad k \equiv 1/r \pmod{2^B}$ because eq (1.1) $(2^S - 3^N)\cdot k=2^B-1$ becomes $ ( - 3^N)\cdot k \equiv -1 \pmod{2^B}$ . (Also it must be smaller than $2^B$)
We get then
$$ (2^B \cdot 2^N - (2^B \cdot n + r))\cdot k = 2^B - 1 \\ 2^B \cdot (2^N - n)\cdot k - 2^B = r\cdot k -1 \\ (2^N - n)\cdot k = {r\cdot k -1\over 2^B } +1 $$ In that equation the lhs is always $lhs \ge k$ but the $rhs \le k$ , so this equation can only hold, if both sides equal k.
3.1) For the lhs this means, that $2^N - n = 2^N - \lfloor 3^N/2^B \rfloor = 1 $
3.2) For the rhs this means, that k must be a divisor of $2^B-1$ and also it must hold $k \le r$. The proof for that property of the rhs was already given in an earlier post .
Condition 3.1) is not further investigated here.
From 3.2) which was the intended part to disprove the nontrivial 1-cycle generally, we can
at least conclude, that the following holds:
then the associated 1-cycle cannot exist.
Unfortunately the last three conditions are not yet formulated as some function of N, so this elementary approach does not yet allow to exclude infinite classes of 1-cycles (for instance it is not even known, whether the number of Mersenne-primes is infinite)