A Possible Proof of Collatz Conjecture

Here’s something that’s been sitting in my inbox for a while: a reader submission claiming to prove one of mathematics’ most notorious unsolved problems.

Eswar Chellappa has been working on the Collatz Conjecture for nearly ten years. He sent me his proof attempt, and I think it deserves a public discussion. Not because I believe it’s correct—I’m skeptical—but because examining why proof attempts fail teaches us more than just accepting “it’s unsolved.”

Let me walk you through his approach, then we’ll discuss the gaps.

What Is the Collatz Conjecture?

The Collatz Conjecture (also called the 3x+1 problem) is deceptively simple:

That’s it. Start with any number. If it’s even, halve it. If it’s odd, triple it and add one. Keep going. The conjecture says you’ll always hit 1.

Example: Start with 7.

$$7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$$

Seventeen steps. But it gets there.

The problem: This has been verified for all numbers up to ( 2^{68} ) (that’s about ( 2.95 \times 10^{20} )). Every single one reaches 1. But nobody has proven it must always happen.

Paul Erdős famously said: “Mathematics may not be ready for such problems.”

Chellappa’s Approach: A Sieve Method

Chellappa’s proof attempt draws inspiration from the Sieve of Eratosthenes, the ancient algorithm for finding primes. Here’s his reasoning:

Setup

Define two helper functions:

• ( f(x) = x/2 ) for even ( x )
• ( g(x) = 3x + 1 ) for odd ( x )

Note that ( g(x) ) always produces an even number (since 3×odd + 1 = even). So after every odd step, an even step must follow.

The Sieve Process

Step 1: Take natural numbers from 1 to 100.

Step 2: Strike out all even numbers (working from 100 down to 2).

Each even number ( x ) maps to ( x/2 ), which is smaller. So:

• 100 → 50 (strike out 100)
• 98 → 49 (strike out 98)
• 50 → 25 (strike out 50)
• … and so on

Step 3: Now only odd numbers (and 1) remain. Since 1 trivially satisfies the conjecture, start from 3.

For each odd number, apply ( g(x) = 3x + 1 ):

• ( g(3) = 10 ) — already struck out, so 3 is “verified”
• ( g(5) = 16 ) — already struck out, so 5 is “verified”
• ( g(7) = 22 ) — already struck out, so 7 is “verified”

Step 4: Continue until ( g(x) > 100 ). This happens at ( x = 33 ) since ( g(33) = 100 ), but ( g(35) = 106 > 100 ).

Step 5: “By suitably increasing the range of numbers, we can conclude that ultimately ( \phi(x) ) must reach 1.”

His Argument Against Loops

The only way the conjecture could fail is if:

1. A sequence goes to infinity, or
2. A sequence enters a loop that doesn’t include 1

Chellappa argues loops are impossible because ( f(x) = x/2 ) is one-to-one. If ( \phi_i(x) = \phi_j(x) \neq 1 ) for ( i \neq j ), we’d have a contradiction.

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The Problems With This Proof

I have to be direct: this is not a valid proof. Here’s why:

Problem 1: Finite Verification ≠ Infinite Proof

The sieve works perfectly for any finite range. But “keep extending the range” is not a proof technique—it’s an observation that the conjecture holds for tested values.

The conjecture has been verified computationally for numbers up to ( 2^{68} ). Chellappa’s sieve method, even extended to 100 million, adds nothing new. We need an argument that covers all natural numbers simultaneously, not a procedure that checks them one by one.

The gap: There’s no logical bridge from “works for 1 to N” to “works for all N.”

Problem 2: The Unbounded Sequence Problem

Chellappa’s loop argument is incomplete. Yes, the function ( f(x) = x/2 ) is injective. But that only rules out immediate repetition.

The real danger isn’t simple loops—it’s the possibility that some sequence grows without bound. A number could, in principle, have a trajectory that increases forever, never returning to previously visited values.

The sieve method assumes all trajectories eventually enter the “already verified” region. That’s exactly what needs to be proven.

Problem 3: Circular Reasoning

The argument essentially says:

1. Every number eventually maps to a smaller number
2. Therefore every sequence reaches 1

But statement (1) is equivalent to the conjecture. You can’t use it as a premise.

Some numbers take many steps before dropping below their starting value. The number 27 requires 111 steps and reaches a maximum of 9,232 before finally descending to 1. The sieve method provides no guarantee that every number eventually descends.

Why Is This Conjecture So Hard?

The Collatz Conjecture resists proof because it combines:

1. Multiplicative and additive operations: The ×3 and +1 interact in chaotic ways
2. Mixing odd/even dynamics: The sequence alternates unpredictably between growth (3x+1) and shrinkage (x/2)
3. No obvious invariant: Standard proof techniques rely on finding a quantity that always increases or decreases. Collatz trajectories do both.

Mathematicians have proven partial results:

• Almost all numbers reach 1 (in a measure-theoretic sense)
• The conjecture holds for all numbers up to ( 2^{68} )
• Any counterexample must be astronomically large

But a complete proof remains elusive.

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What Would a Real Proof Look Like?

A valid proof would need to establish one of these:

1. A global descent argument: Show that some measure (not just the number itself) strictly decreases on average, guaranteeing eventual arrival at 1.
2. Structural analysis: Prove that the “tree” of all trajectories leading to 1 covers all natural numbers.
3. Impossibility of counterexamples: Prove that divergent sequences and non-trivial loops are structurally impossible.

Terence Tao (2019) made progress on approach #1, proving that “almost all” Collatz orbits attain almost bounded values. But “almost all” isn’t “all.”

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A Note on Attempted Proofs

I want to be clear: I’m not mocking Chellappa’s effort. Working on hard problems is how mathematics advances. Most proof attempts fail, and that’s fine.

The Collatz Conjecture has attracted thousands of attempted proofs. All have failed so far. The pattern of failure is instructive:

• Most assume what they’re trying to prove
• Many confuse empirical verification with logical proof
• Some find genuine partial results but can’t complete them

If you’re working on this problem, my advice: focus on understanding why previous attempts failed before constructing your own. The Math Overflow archives are full of instructive examples.

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