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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.

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closed as off topic by lhf, Raskolnikov, GEdgar, Hans Lundmark, Steve D Sep 27 '12 at 22:09

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Based on meta thread Submitting a paper for review I don't think this type of question is suitable for this site. –  Martin Sleziak Sep 27 '12 at 6:17
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Awesome! Any chance you could suggest a forum appropriate for such content? –  x1101011x Sep 27 '12 at 6:38
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@x1101011x IMHO there should be doubt in your mind that you've proven it; considering how long the problem has been open, skepticism of such a straightforward proof is not only natural but arguably healthy, and that includes from the proof's author. That said, I do feel like the proof is worth much closer examination; it's not obviously wrong. My biggest immediate concern is that it seems like the mechanisms in 3.5-3.6 should provide a bound on the maximum number of iterations, and a relatively small (and thus possibly falsifiable) one at that. –  Steven Stadnicki Sep 27 '12 at 7:41
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I'm skeptical for the simple reason that you start by writing that you have a "proof of the Collatz problem." One can prove a theorem; one can solve a problem; one can't prove a problem, and anyone who writes such a thing.... –  Gerry Myerson Sep 27 '12 at 12:44
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And we can now see why such questions are not liked here. It becomes a discussion. So it should be posted on a discussion site. –  GEdgar Sep 27 '12 at 18:04

2 Answers 2

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.]

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Actually, if you look more closely at theorem 3.6, I think you'll find that it shows we eventually reach a number such that all numbers after it are congruent to $1$ modulo $8$, which means we'll eventually reach a point where it will decrease until $1$ and stay there. –  x1101011x Sep 27 '12 at 13:54
    
Let me give you a constructive example: Say you want to find all numbers that go up once. Then, the algorithm gives you the form $4x-1$, and you can see that $T(4x-1)=ax-b=6x-1$, which means $3,7,11,15,19,\ldots$ map to $5,11,17,23,29,\ldots$, respectively -- obviously true. Now, if you want to extend your path, you have three choices: up, up; up, down; up, down a lot, i.e., $S=3,3$, $S=3,1$, or $S=3,5$. However, by the table in lemma 3.5 (row $(6,1)$), you see that these numbers have the forms $8x-1$, $16x-5$, and $16x-13$, or the minimums are up, up: $7$, up, down: $11$, up, down a lot: $3$. –  x1101011x Sep 27 '12 at 14:12
    
As you can see, only in one case did the minimum value stay the same. So, if we were to start with the number $7$, then, knowing it goes up, up and that the minimum number that goes up, up is $7$, we now know that this number can only follow the bold path in our table. But this bold path quickly ends in every number being congruent to $1 \pmod{8}$. This is how we know the Collatz theorem. Do you see it? –  x1101011x Sep 27 '12 at 14:15

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.

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I'm not sure what you are trying to say. The transformation $T$ is defined on all odd numbers. Could you elaborate on how exactly you find the proof "much too short a thought?" –  x1101011x Sep 27 '12 at 14:00
    
@x1101011x : The reason for my concern is the prominence that I've seen for the modularity wrt 8. I've not yet worked through your paper - perhaps I can go into it more seriously in the evening. Possibly we should also switch to email-conversation in case the problems should be trickier and need more discussion/explanation (my email adress is in my profile) –  Gottfried Helms Sep 27 '12 at 14:12

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