Implied bound falls short of Baker
The argument
Because the separation between powers of 2 and 3 that the Collatz conjecture implies is far weaker than what Baker's theorem gives, a proof of Collatz would yield only a weaker transcendence-type statement, not Baker's theorem itself, so the comparison holds at most in the loose sense that a Collatz proof must produce some nontrivial transcendence result.
Does it work?
Not yet evaluated.
Premises
- UnassessedNo current assessment. Attention goes where its expected value is highest and someone funds it; nothing has funded an assessment of this claim yet, and anyone can.constitution →The lower bound on |2^n − 3^m| implied by the weak Collatz conjecture is far weaker than the bound Baker's theorem gives.
Where it is made
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Earlier ratings
- Sep 16, 2026
Granting that the implied separation is far weaker than Baker's bound, which Tao himself states, it follows that a proof of Collatz would not by itself reprove Baker's theorem, so the literal comparison is not established. The argument only qualifies the claim: it leaves untouched the weaker and defensible reading that any Collatz proof must produce nontrivial transcendence-type results.