Collatz Conjecture proof that no cycle can exist other than the 1,4,2 cycle. Can someone verify it?

Question

This is not a proof of the Collatz Conjecture, but I somehow managed to show that there is no cycle that can exist other that the 1, 4, 2 cycle:

$n$ is a positive integer

$3n+1$, $n$ is odd

$\frac n2$, $n$ is even

First, let us assume that a cycle that does not contain 1 exists. This cycle must contain an odd number within it, otherwise it would be possible to continously divide each number in the cycle by 2, which would lead to 1. Thus, let us begin this cycle with that odd number and because this is a cycle, it does not matter where we begin. Let that odd number be $A$, where $A$ is not equal to 1. Since $A$ is odd, we apply the odd equation to find the next term: $3A+1=B$. $B$ must be an even number since, $3A$ will produce an odd number and adding $1$ to an odd number must give an even number. To find the next term we apply the even formula to $B$: $\frac B2=C$.

Now becuase the domain of $n$ is defined to be the positive integers, $n<3n+1$ must be true for all $n$, thus $A<B$. By the same logic $3n+1>\frac{3n+1}{2}$ thus $B>C$. Now $n<\frac{3n+1}{2}$ only when $n>-1$, so becuase the domain of $n$ is all positive integers we can say that $A<C$. Thus we can form the inequality $A<C<B$.

Since this is a cycle, the values within it must return to $A$ at some point. Let us denote the value right before $A$ by the letter $Z$. Since $A$ is an odd number, $Z$ must be even because the only equation that can return an odd number is $\frac n2$. Becuase $\frac Z2=A$, $Z<A$ must be true. Because $Z<A<C$ is true, $Z<C$ must also be true.

Becuase $\frac Z2=A$, we can say that $Z=2A$. We also know that $3A+1=B$ so $A= \frac{B-1}{3}$. Substituting for $A$ we get $Z=2(\frac{B-1}{3})$. Again, we also know that $\frac B2=C$ so $B=2C$. Substituting in for $B$ we get $Z=2(\frac{2C-1}{3})$ or $Z=\frac{4C-2}{3}$.

Because we know $Z<C$, we can say $\frac{4C-2}{3}<C$ which is only true when $C<2$. The only positive integer less than 2 is 1, so $C=1$. Because $\frac B2=C$, $B=2$. Also because $3A+1=B$, $A=\frac 13$. The domain of $n$ is all positive integers, thus this is a contradiction. Therefore, there exists no cycle that does not contain the number 1 given the parameters set by the conjecture.

Answer 1

"Because $\frac{Z}{2}=A \Rightarrow Z \lt A$"

Is not true. However,

$$\frac{Z}{2}=A \Rightarrow Z \gt A$$

If $Z \lt C$ then your argument is unaffected.

If $Z \gt C$ then your argument is affected. Specifically,

$$\cfrac{4C-2}{3} \not \lt C$$

But rather,

$$C \lt \cfrac{4C-2}{3}$$ $$\Rightarrow C \gt 2$$

Answer 2

"Because $\frac{Z}{2}=A, Z<A$ must be true. "

Your contradiction is a consequence of this inequality, which is wrong.

Answer 3

A further, and more fundamental oversight which you may want to look at is that I think your proof doesn't cater for repeated sequences of dividing by two in-between each $3x+1$. A powerful way to attack the existence of nontrivial loops in the Collatz conjecture is to start with the minimal odd number in some cycle and then to attempt to prove that it is 1, or that if it's not 1 then there is necessarily some lower odd number in the cycle.

Answer 4

You assume $C=B/2$ and then argue for relation of $A$ and $C$ only. But this means, you are only looking for a 1-odd-step cycle (other than the trivial one). But in half of the cases $C=B/4$ or $C=B/8$ and so on. Moreover, even if $C=B/2$ and $C$ is odd, then further steps with $D,E,F,G,...$ might follow, in which the trajectory goes back near to or below $A$. So not much, if anything at all, has been proved there...

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The proof for the (odd) 1-step-cycle (which I understood you have attempted, hope I didn't misunderstood you) is easy: $$ A= {3A+1 \over 2^a} \qquad \quad \text{with } a\ge 1 \text{ such that the rhs is odd}$$ Then $$ 2^a = {3A+1\over A} \implies 2^a = 3+\frac1A $$ and the only possible positive integer on the rhs is $4$ which is also $4=2^2$ so $a=2$ follows; with this we have the solution $A=1$ and no other solution (for $A \ge 1$) is possible.