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Mike Johnson
Mike Johnson
May 15, 2026

Why is the Collatz conjecture still hard to prove?

The Collatz conjecture: Start with any positive integer nnn.

If nnn is even, divide by 2: n→n/2n \to n/2n→n/2.
If nnn is odd, multiply by 3 and add 1: n→3n+1n \to 3n + 1n→3n+1.

Repeat. The conjecture is that you ALWAYS reach 1 eventually.

Example starting with 7:

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

It took 16 steps! And it went all the way up to 52 before coming down.

Verified for all numbers up to 2682^{68}268 but NO proof exists.

Whats makes this so hard? It seems like it should be easy to prove by induction or something. Why do mathematicians say this problem is "dangerous" and can consume years of your life?

Are there ANY partial results? Like does it hold for almost all numbers in some statistical sense?

1 answers4.3k views

1 Answer

-4
Chen Wei
Chen Wei
May 16, 2026
Accepted
The Collatz conjecture is hard because it involves both multiplication and addition in a way that destroys algebraic structure. **Why induction fails:** In standard induction, you prove $P(n)$ for base case and show $P(k) \implies P(k+1)$. But Collatz isnt monotonic — $n$ might go up before going down. Theres no obvious well-ordering to induct on. **Why it resists pattern analysis:** The $3n+1$ operation sends numbers up, the $n/2$ operation sends them down. Whether a number eventually reaches 1 depends on the delicate balance between these operations across all numbers. This is a dynamical system with sensitive dependence on initial conditions. **Partial results:** - Terras (1976): Almost all numbers eventually reach a value less than the starting number ("almost all" in the sense of natural density 1) - All numbers up to $2^{68}$ have been verified computationally - It can be shown that any counterexample must be extremely large (more than about $10^{20}$) **Why mathematicians say its dangerous:** People (often amateur mathematicians) get obsessed with proving it, spend years on it, and produce flawed proofs. Erdos said "mathematics is not ready for such problems." **The function $T(n)$ (number of steps for $2^n - 1$):** Yes, there are known formulas for specific families. Numbers of the form $2^k - 1$ have a predictable growth pattern because in binary theyre all 1s, and the $3n+1$ operation on a binary number of all 1s has interesting properties.
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