Our reader Eswar Chellappa has sent his work on the solution of ‘3X+1’ problem, also called Collatz Conjecture. He had been working on the proof of Collatz Conjecture off and on for almost ten years. The Collatz Conjecture can be quoted as follow:

Let $\phi : \mathbb{N} \to \mathbb{N}^+$ be a function defined  such that:
$$\phi(x):= \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even } \\ 3x+1, & \text{ if } x \text{ is odd} \end{cases}$$

Then the iterates of $\phi(x)$ will eventually reach $1$ for any initial value of $x$.

See this post about Collatz Conjecture for more details.

Plenty of proof attempts were made by various mathematicians. But none of those could flawlessly prove the statement. Mr. Chellappa’s attempt is based upon the famous Sieve of Eratosthenes. Despite of his experience & confidence, I can not guarantee if this work is perfect. I invite readers to cross check the flawlessness and tell what they think.

## Introduction

Let, $f(x) = x/2$ if $x$ is even and  $g(x) = 3x + 1$ if $x$ is odd. Since $3x+1$ is an even number for any odd $x$, we can replace any odd number by an even number which equals to $3x+1$. And when, $3x+1$ is an even number, we can successfully halve it according to first step of the function defined in the conjecture.

## The method proposed below is similar to the famous Sieve of Eratosthenes

To start with the proof, choose the natural numbers from 1 to 100.

Since we have $$f(x) = x/2$$ for even $x$. Let’s strike out all even numbers for each of them reduces to another odd or even number (less than x) by f(x), by exactly one iteration. This we do from 100 to 2, in the decreasing order.

• We strike out 100 as $f(x)$ reduces it to 50.
• Then we strike out 98 keeping 49 as the result of one iteration.
• 50 is also struck out giving 25 as result of one iteration.

Now all it remains, the odd numbers and natural number $1$. As $1$ automatically verifies the conjecture, we start from 3 and end to 99.

• We have $g(3) := 10$. Since $10$ has already been struck out, we can simply say that $3$ satisfies the conjecture. Hence we strike out 3.
• $g(5) := 16$, so 5 can also be struck out.

By this process we can strike out all odd numbers up to 33 as we have taken only numbers up to 100 and $g(35) >100$.

By suitably increasing the range of numbers we can conclude that ultimately $\phi (x)$ has to reach 1 irrespective of the starting natural number $x$.

## Any possible worries?

The only possibility that might worry us was falling in to a loop. This happens only if $\phi_i(x) = \phi_j(x) \neq 1$ where $i\neq j$. But that is not possible since $\phi (x)=x/2$ is one to one.

## Remark

• By striking out the numbers, we verify that they have images under $f(x)$. That is to say, more iterations are possible until the sequence reaches $1$.
• Concerned about $\mathbb{N}^+$ notation?

PS: This isn’t a paper.

A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that.

1. This isn’t remotely a proof. Everything you wrote also applies word-for-word to the function $f(x) = x/2$ if x is even and $f(x) = 5x+1$ (instead of 3x+1) if x is odd. But the conjecture is false for this different function, since it contains the following loop: 13, 66, 33, 166, 83, 416, 208, 104, 52, 26, 13

In other words: you barely wrote a single thing that involves more than the fact that f sends even numbers to odd numbers and odd numbers to even numbers, and no such argument is going to be enough to prove the Collatz conjecture.

2. Methyboy states precisely why this is not even remotely a proof.

Let’s give an example based on the famous Collatz-number 27.

Is 54 safe because it reduces to 27, which amounts to 82 only, which reduces again to a smaller number 41? Not at all, since 41 amounts to 124, which is outside the range 0..100. Since 3×41+1 exceeds 100, the proof does not apply to 41, but since since 41 can be reached from 54, the proof does not apply to 54 either.

3. I was expecting to see some attempts using mathematical induction which restricts the case that disporves the conjucture to a number of the form 4n+3.
Progressing through 1,2,3,4,5 etc. which we know support the conjecture, if we assume there is a number x that defies the conjecture, that number has to be of the form 4n+3. Because , 4n and 4n+2 will reduce to an already passed number. 4n + 1 will become (12n + 4)/2 = 6n + 2 in the next step and 3n+1 in next., again a number aready passed. Since the cycles of random length, it cannoot be supposed that 4n+3 will get reduced to diffent number mod 4 or a product of a suitable power of 2 in a predictable number of steps. Eager to learn about attempts in this route if any.

• Babu, you at least provide us with some notions of general applicability for attacking the problem, some of which it should be fairly easy to state and prove as lemmas. By collecting a group of lemmas of sufficient scope and power, we should then be able to reduce the problem cases to a handful or less.

The sieve notion mentioned by the OP, if combined with mathematical induction as you suggest, could perhaps provide one such reduction or simplification of the problem domain. I invite all readers to try combining them to good effect!

Elsewhere, I’ve seen statements to the effect that the most problematic cases all involve infinite cycles. (How hard that would be to show, I haven’t yet tried to ascertain. There’s a nice little project – best done oneself, for the benefit of the exercise! – before checking out others’ work.)

In such a case, there would exist integers n>1 and x>1, such that ϕ^n(x)=x, but also, for all 1<i<n, ϕ^i(x)>1. Obviously, if we could show that this never happens, it would be a great step forward. Equally obviously, showing that won’t be easy, or it would have been done already by one or more of the thousands who’ve cracked their brains over the Collatz Conjecture. However, it’s worth thinking about how to break this problem down into different cases; for example, would any such n necessarily be even?