Is this proof for no non-trivial Collatz cycles flawed?

Question

I've been working with the Collatz Conjecture a lot lately as a way to distract me from the math I'm supposed to be doing for school. Anyway I feel like I've constructed (not very rigorously I might add) a proof that states that the 4-2-1 cycle is the only cycle under iteration of the Collatz function. Not much else to say but numbers! Here it is:

We know that $$3(2a_n+1)+1=2^{P_{n+1}}(2a_{n+1}+1)$$ This is the Collatz-iteration function, with each division step $(\frac{x}{2})$ written into each multiplication step $(3x+1)$. Instead of $x$, I'm using $2a_n+1$ to ensure that an odd value is being placed into the multiplication step. This doesn't inherently lose any information about what odd numbers come up in a given Collatz sequence.

Normally, we have $$a_n\in\Bbb{N^0},a_n\ge0$$ However, if a cycle is going to exist that isn't the trivial cycle, it's not going to have $1$ as an odd number in the sequence, as that would it the trivial cycle. Imposing this restriction, the last line changes slightly: $$a_n\in\Bbb{N},a_n\gt0$$ This makes any $2a_n+1\gt1$, specifically $2a_n+1\in\{3,5,7,9,...\}$. Hold onto that, this is a critical assumption we are making about the $a_n$'s in this non-trivial cycle.

---

If there is a cycle of length $N$, the first equation holds for all $a_n,n\lt{N}$ (even if it isn't a cycle), giving $$3(2a_1+1)+1=2^{P_{2}}(2a_{2}+1)$$ $$3(2a_2+1)+1=2^{P_{3}}(2a_{3}+1)$$ $$...$$ $$3(2a_{N-2}+1)+1=2^{P_{N-1}}(2a_{N-1}+1)$$ $$3(2a_{N-1}+1)+1=2^{P_{N}}(2a_{N}+1)$$ What makes it a cycle of length N is this very last equation: $$3(2a_N+1)+1=2^{P_{1}}(2a_{1}+1)$$That is, the last number returns to the first number in the cycle.

Rearranging a typical equation in the sequence, $$3(2a_n+1)+1=2^{P_{n+1}}(2a_{n+1}+1)\Rightarrow \frac{3(2a_n+1)+1}{2a_{n+1}+1}=2^{P{n+1}}$$ Now, start the magic and multiply all these equations together, that is: $$\prod^N_{n=1}\left[\frac{3(2a_n+1)+1}{2a_{n+1}+1}=2^{P{n+1}}\right]$$ where $a_{N+1}=a_1$

This now gives $$\prod^N_{n=1}\left[\frac{3(2a_n+1)+1}{2a_{n}+1}\right]=2^{\sum^N_{n=1}P_{n}}$$

I changed the index here! I thought this might be where the mistake lies but I'm not sure. It seems plausible to do as each term is being multiplied together, so by the commutative law of multiplication I should be able to move those products around so that the index is neater. Anyway...

Taking the $\log_2$ yields $$\sum^N_{n=1}\left[\log_2\frac{3(2a_n+1)+1}{2a_{n}+1}\right]={\sum^N_{n=1}P_{n}}$$

Reducing,$$\sum^N_{n=1}\left[\log_2\left(3+\frac{1}{2a_{n}+1}\right)\right]={\sum^N_{n=1}P_{n}}$$

Knowing that $P_n\in\Bbb{N}$, we know that $\sum^N_{n=1}P_n\in\Bbb{N}$ as well, further implying that $\sum^N_{n=1}\left[\log_2\left(3+\frac{1}{2a_{n}+1}\right)\right]\in\Bbb{N}$, even further implying that $\log_2\left(3+\frac{1}{2a_n+1}\right)\in\Bbb{N}$. However, this restricts $a_n$ to $$\frac{1}{2a_n+1}\in2^\Bbb{N}-3,\frac{1}{2a_n+1}\in\{-2,-1,1,5,13,...\}$$

Referring to our assumption earlier that $a_n\in\Bbb{N},a_n\gt0$, we know that $$\frac{1}{2a_n+1}\in\{\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},...\}$$ thus contradicting our conclusion of $$\frac{1}{2a_n+1}\in\{-2,-1,1,5,13,...\}$$

Therefore to have a cycle of length $N$,$$a_n=0,2a_n+1=1$$ must be one of the $N$ terms, thus making it the trivial cycle. QED.

What happened??

---

Also important to note, if $a_n=0$ is allowed, then the two sets constructed from $a_n$ share the odd number 1, the trivial cycle. Just thought that extra bit was interesting.

Answer 1

Your inference from the sum of the logarithms being a natural number to each logarithm being a natural number is invalid. As Maximilian Janisch noted in a comment, a good heuristic to test your proof is to apply it to the map $n\mapsto3n-1$. In this case there is a non-trivial cycle $5,10,20,7,14$. Applying your proof to this case leads to the sum of logarithms

$$ \log_2\left(3-\frac15\right)+\log_2\left(3-\frac17\right)=3\;, $$

which is indeed the case since $\left(3-\frac15\right)\left(3-\frac17\right)=8$, but this doesn’t imply that each factor is a power of $2$ (and thus its binary logarithm a natural number).

Answer 2

The gap in the argument is here:

Knowing that $P_n\in\Bbb{N}$, we know that $\sum^N_{n=1}P_n\in\Bbb{N}$ as well, further implying that $\sum^N_{n=1}\left[\log_2\left(3+\frac{1}{2a_{n}+1}\right)\right]\in\Bbb{N}$, even further implying that $\log_2\left(3+\frac{1}{2a_n+1}\right)\in\Bbb{N}$.

Even if the sum is an integer, it does not necessarily imply that all the individual terms are integers.

This is somewhat related to the relabeling of indices in the denominators of the product. Taking the logarithm before relabeling, we get the sum $$\sum^N_{n=1}\left[\log_2\left(\frac{3(2a_n+1)+1}{2a_{n+1}+1}\right)\right] =\sum^N_{n=1}\left[\log_2\left(3(2a_n+1)+1\right)-\log_2\left(2a_{n+1}+1\right)\right].$$ To condense notation a bit, denote by $$A_n:= \log_2\left(3(2a_n+1)+1\right)\quad\text{and}\quad B_{n+1}:= \log_2\left(2a_{n+1}+1\right)$$ the terms appearing in the sum. The relabeling of denominators corresponds to regrouping the summands as \begin{align*} \sum^N_{n=1}\left[A_n-B_{n+1}\right] &=[A_1-B_2] + [A_2-B_3] + [A_3-B_4] + \dots + [A_{N-1}-B_N] + [A_N-B_1] \\&=A_1 + [-B_2+A_2] + [-B_3+A_3] + \dots + [-B_N+A_N] - B_1 \\&=\sum^N_{n=1}\left[A_n-B_{n}\right] \end{align*}

In the original sum each individual term $$A_n-B_{n+1}=\log_2(2^{P_{n+1}})=P_{n+1}$$ is indeed an integer, but there is in principle no reason for the terms $A_n-B_n$ of the reorganized sum to be integers.