MethodMath
Mike Johnson
May 15, 2026

Why is the collatz conjecture so hard to prove when it seems so simple

The Collatz conjecture: Start with any positive integer nn.

If nn is even, divide by 2: nn/2n \to n/2.
If nn is odd, multiply by 3 and add 1: n3n+1n \to 3n + 1.

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

Example starting with 7:

72211341752261340201051684217 \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

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

Verified for all numbers up to 2682^{68} 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 answers3.4k views
4 comments
Mike Johnson
Mike JohnsonMay 16, 2026

i spent 3 hours playing with this after reading and got nothing useful. this problem is cursed.

Alex Kim
Alex KimMay 16, 2026

theres a 5000 dollar prize for anyone who proves it. but trust me, its not worth the years of frustration.

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1 Answer

Prof. Chen Wei
Prof. Chen WeiMay 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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