Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer?

Question

For me, one of the most interesting results of the Collatz Conjecture is that if an $m$-cycle exists, then $m > 68$. So, there are no non-trivial $1$-cycles, $2$-cycles, ..., up to $68$-cycles. The details behind this argument can be found in this paper by Simons & De Weger (2003).

Let:

• $\nu_2(x)$ be the 2-adic valuation of $x$.
• $u = 3^kt - 1$
• $b = \nu_2(u)$
• $a = \nu_2(\frac{u}{2^b}+1)$

If a $1$-cycle exists, then the maximum value of the cycle is equal to $3^kt - 1$ where $k,t$ are positive integer and the lowest value of the cycle is equal to $2^kt - 1$ characterized by the following:

$$u = \left(\frac{3}{2}\right)^a\left(\frac{u}{2^b} + 1\right)-1$$

So that:

$$\left(\frac{2}{3}\right)^a(u+1) = \frac{u}{2^b} + 1$$

And:

$$\frac{u(2^{a+b} - 3^a)}{3^a2^b} = \frac{3^a - 2^a}{3^a}$$

Which reduces to:

$$\frac{u}{2^b} = \frac{3^a - 2^a}{2^{a+b} - 3^a}$$

It seems to me that since no non-trivial $1$-cycle exists, it follows that if $a,b$ are positive integers where $ab > 1$, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not a positive integer.

Here's the argument:

(1) Assume that there exists positive integers $a,b$ with $ab > 1$ such that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is an integer.

(2) Let $x = \frac{3^a - 2^a}{2^{a+b} - 3^a}$

(3) Let $u = 2^bx$

(4) Let $k = a$, $t = \left(\frac{1}{2^a}\right)\left(\frac{u}{2^b}+1\right)$ which are both integers.

(5) So it follows that $u = 3^kt - 1$

(6) But then we have a contradiction since no $1$-cycle exists. Therefore, we can reject our assumption at step(1).

Is this argument valid? Does the result from Simons & De Weger (2003) prove that with the given assumptions, $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ is not an integer?

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Edit: I added the clarification that $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be a positive integer.

@rukhin makes a great point. For positive integers $a=2$ and $b=1$, $\dfrac{3^a - 2^a}{2^{a+b} - 3^a} = -5$

This is correlated with the negative $1$-cycle with $k=1$ and $t=\left(\frac{1}{4}\right)\left(\frac{-10}{2} + 1\right) = -1$

The standard collatz sequence defined as:

• $T(x) = \begin{cases} \frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\ \frac{1}{2}x, && \text{if }x\text{ is even}\\ \end{cases}$
• $T_1(x) = T(x)$
• $T_{c+1}(x) = T(T_c(x))$

So that:

$$-1, -2, -1, \dots$$

Answer 1

Say you start with $n_0=2^kt-1$, apply $a$ times the Collatz function $f(n)=\frac{3n+1}{2}$, than $b$ times the Collatz function $f(n)=\frac{n}{2}$, you can easily see that you end up with $$n=\frac{3^a}{2^{a+b}}n_0+\frac{3^a-2^a}{2^{a+b}}$$ or $$n\cdot{2^{a+b}}-3^a\cdot n_0=3^a-2^a$$ and in the case of a cycle $n=n_0$ $$n=\frac{3^a-2^a}{2^{a+b}-3^a}$$ Note: $a=k$, the value $a = \nu_2(\frac{u}{2^b}+1)$ you gave is for the next iteration

Now in case of a cycle, it has been shown that (you can find it by noticing that $\frac{1}{2}<\frac{3^a}{2^{a+b}}<1$): $$a+b=\lceil a\cdot \log_2(3)\rceil $$ So indeed, if there is such an integer $n=\frac{3^a-2^a}{2^{a+b}-3^a}$, it must be part of a 1-cycle, and you can conclude there is none (outside the trivial one), but it would only be valid within the above constraint on $a+b$.

Also notice that the question has already been raised.....by you Is this argument valid for showing given integers $n>1$ and $2^m > 3^n$, $(2^m - 3^n) \nmid 3^n - 2^n$

Also note that a m-cycle does not mean m successive 1-cycles, so there is no such constraint on $a+b$ within the m-cycle (but in this case you don't have $n=n_0$ or the equation you mentioned)