Welp, I proved the Collatz Conjecture

Read properly

Thanks for the free 500

and left as an exercise to the reader

So you mean: (2^k - 1)/3 is composite implies that it is also an even power of 2?

no, it's just a multiple of it
I'm not OP but I understand his proof now. it was all so fucking obvious

Is it an actual proof??

no

OMG if you prove the conjecture those yid give you only 500$ ? is this real

What's a multiple of what? (2^6 - 1)/3 = 21 isn't a multiple of any power of 2

i don't know what's considered a 'formal' proof, but if it's not a proof it's a much stronger form of the collatz conjecture

still checking it

>posted 3 hours and 10 minutes ago
>nobody has yet spotted a flaw
welp, Veeky Forums solved a millenary problem

from Lemma 4 onwards it's all handwavy nonsense. there's no flaw because there's no proof

what he means is (2^(2n))-1 is always divisible by 3. Or rather, 4^n -1 is always divisible by 3.

Another way to think about it is 4^n = (3+1)^n

Oh, that's it? That's quite trivial. OP's wording was terrible.

no you didnt

Don't be perelman fag.
You better have posted this first on arxiv.

If not then someone call Mochizuki or a mod to tell him to save the IP of this bastard and them track him down to give him his prize. Motherfucker.

are you being ironic?

are you stupid? this is a big deal
i'm angry because i really wanted the 500 dollars, but that won't make me pretend this isn't a big deal

again, are you being post-ironic or something? this isn't a proof

Good goy. Recognition and money are for other people. You should just do work for free.

Fuck no.

If this proof is true then he better receive his prize. Stop being a faggot.

Prove him wrong or call JapMoot to save his bastard's IP.

CALL THE FUCKING FBI AND NSA. WE ARE TRACKING THIS FAGGOT TO GIVE HIM 500 BUCKS.

>delusional retard thinks he's funny

no wonder you have no friends user, nobody can tolerate this kind of shit in real life.

Posted 5 hours ago

Nobody has debunked it

Logic holds

This is it gentlemen

Nah, it doesn't work I've tried this before. You're basically looking in mod 4 and saying that 0, 1, and 2 all converge under the iteration of the rules.

But numbers congruent to 3 mod 4 don't necessarily converge since after 3n+1 you increase but are not guaranteed to decrease sorry.

I already covered that. Read carefully.

How does this prove the non-existence of other loops that may exist? It seems you just proved that you can reverse the Collatz on any odd number, but not that all odd numbers lead to one with the Collatz rules, or that all odd numbers can be derived from one using the reverse Collatz. I could just be retarded.

Erdős is a greedy guy
>Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[9] He also offered $500 for its solution

Replies 35
Unique Posters 17

Something doesn't ADD up here.

Posts with inline quotes: 25

It's almost like people exchange two messages

Why does he write like it's the 1500s and we haven't developed math notation?

I don't know shit about math. Is this legit?

no you didn't. everything after "Lemma 4" is nonsense

Holy shit, could you define what symbols mean before you use them? This is bordering on unreadable

that's the least controversial part... the notation is fucking standard

Agreed.

wtf is the main idea of this proof?

Why would I be expected to know what fucking [math[\/mu[/math] means? You don't get to say notation is standard in math. You have to define your symbols so that others can read it.

I read this interesting answer about why we care about the Collatz conjecture and its generalizations.

First answer here: math.stackexchange.com/questions/2694/what-is-the-importance-of-the-collatz-conjecture

That's a really interesting take on the ideas underlying the conjecture. I wonder if it's possible to develop some general notions along these sort of "conservation of prime energy" lines that could ultimately lead to a tidy proof.

>Zooms in
>looks at a link on the 3rd page
Nice try faggot

I don't even get lemma 1 D;

You don't get lemma 1? Most of the first page is dedicated to saying that 2^k \equiv 1\pmod{3} iff k is even, and that 2^k+2^{k+1}\equiv 0\pmod{3}, but that is obvious since 2\equiv (-1)\pmod{3} and the most basic understanding of modular arithmetic. The statement of lemma 2 is badly worded (and false as written with counterexample k=4 as already pointed out). The word composite should be replaced with "is an integer."

It is from there that the handwaviness begins with the OP introducing \mu without properly defining it. Even so, it seems he is trying to build the list of what numbers will reach 1. He notes that for those numbers which do reach one, the final set of moves will be repeated divisions by two or will be multiplying by three and adding one followed by repeated divisions by two. Again, nothing new or surprising there.

From there, he tries to use some function f^a which he has not defined or explained what it is. Without having told us what f is, everything from here out is meaningless.

this is an accurate and well-articulated summary of the flaws in the OP's proof

bump

this is history happen to today in the eye

That's quite much for something this irrelevant.

>don't believe this
>still posting to be part of history just in case

I'm here for the memes

f is defined right there, dumb shit

it's defined by parts

it's a picture from google images

it's unrelated to the proof

kek