What to do with Collatz proof? [closed]

Question

I have discovered a proof solving the Collatz problem, and I have no idea what to do with it. Given the nature of the topic, all the experts I've found that are capable of reviewing the paper instantly dismiss it.

However, I have the pressing need to share it. It's so simple and so clear, though I'm sure I haven't done the proof justification with my presentation of it. Please, take the time to look through this proof, and ask for clarification on any unclear parts. It's worth it.

Be extra critical. All corrections are welcome.

Answer 1

The majority of the paper seems unused in the proof of the Collatz Conjecture. Your proof appears to be:

• Step 1 (Theorem 3.6): For any $x \in \mathbb{Z}^+$, if we apply the Collatz transformation repeatedly $x,T(x),T^2(x),\ldots$, we eventually reach an element congruent to $1 \pmod 8$.

• Step 2 (Proposition 3.2): If $x \equiv 1 \pmod 8$, then $x \mapsto 3x+1 \mapsto (3x+1)/2 \mapsto (3x+1)/4$ where $(3x+1)/4 < x$ provided $x>1$. This is also true.

• Step 3: Repeatedly apply Step 2 until you hit $1$.

However, even if $x \equiv 1 \pmod 8$, it doesn't imply $(3x+1)/4 \equiv 1 \pmod 8$. E.g. $(3 \times 9+1)/4=7 \equiv -1 \pmod 8$. So, we can't repeatedly apply Step 2.

[Correction: I originally wrote "This can be verified via elementary number theory" in regards to Step 1, but I no longer think this is true, either. For example, starting from $433$, we get the sequence (mod 8):

1, 4, 2, 5, 0, 0, 4, 2, 5, 0, 4, 6, 7, 6, 3, 2, 5, 0, 0, 0, 4, 2, 5, 0, 0, 4, 2, 1

in which we don't hit $1 \pmod 8$ until right at the end (when we reach 1 itself). But, for the sake of argument, let's assume Step 1 has been proved.]

Answer 2

Since you try to make something by expanding the division-criteria with respect to moduli by higher powers of 2, you might be interested in this: Consider the table of transfers depending on modularity by some power of 2 for numbers of the form $2^Ak+r$ where k is some nonnegative integer
$ \begin{array} {l|l|l} j&\text{odd n of the form} & \text{odd n of the form} & \\ \hline 1& 2^2 k + 3 & 2^3 k +1& \\ 2 & 2^4 k + 13 & 2^5 k +5& \\ 3& 2^6 k + 53 & 2^7 k +21& \\ \cdots& \cdots & \cdots & \\ j& 2^{2j}k + {10\cdot 4^{j-1}-1\over4-1} & 2^{2j+1} + { 4^j-1\over4-1}\\ \hline\\ result &\to 6k+5 & \to 6k+1 \end{array}$
then, as I understood your paper, you're concerned with numbers of the form in the first two rows only. However, any number of the forms in the upper part of the table result after one transformation either to the form 6k+5 or 6k+1 and this latter forms can again mean any form in the upper part of the table (where then another k, one may indicate it by another letter, say j, must be written).

The upper part of the table covers the whole set of odd integers (without overlap) so I think that your observation concerning that (mod 8) and then the conclusion is much too short a thought.