Proving the Collatz conjecture is at least as difficult as proving Baker's theorem on linear forms in logarithms.
Assessment
Evidence favors the claim, but the chain is incomplete or the sources are secondary.
The comparison originates with Terence Tao and rests on a rigorous observation rather than a mere impression of difficulty. The Collatz conjecture includes, as a weak consequence, the statement that the only cycle of the Collatz map on the positive integers is the trivial one, and by an observation of Böhm and Sontacchi that statement is equivalent to the insolubility of a divisibility problem in powers of 2 and 3. Tao showed in 2011 that the absence of nontrivial Collatz cycles already implies a nontrivial lower bound on the gap between 2^n and 3^m: if the gap were too small, an additive-combinatorial argument (a large parallelepiped in a small cyclic group) would manufacture a cycle. Since every known nontrivial lower bound of this kind comes from transcendence theory, in Gelfond's and Baker's work on linear forms in logarithms, any proof of the Collatz conjecture would have to either invoke such results or supply a new method capable of producing them. This is what Tao means by saying the conjecture is "basically" at least as difficult as Baker's theorem, and it is corroborated by the fact that the partial results excluding structured Collatz cycles all run through Baker's theorem.
Taken literally, the comparison overstates what has been shown. The gap that the Collatz conjecture implies is far weaker than the exponential separation Baker's theorem provides; Tao himself describes his conditional bounds as very weak by comparison. A proof of Collatz would therefore yield a nontrivial transcendence-type result, not Baker's theorem itself, and a later expository account that says a Collatz proof "would necessarily yield a proof of Baker's Theorem" goes beyond the argument it cites. The defensible content of the claim is Tao's own careful version: any proof of the Collatz conjecture must use existing transcendence theory or contribute new methods to it, which already rules out many proposed elementary approaches. Read that way the claim is well grounded and undisputed in the literature; read as a strict statement that Collatz implies Baker's theorem, it is not established.
Full reasoning: the evidence and decisions behind this verdict
The claim is a judgment about proof difficulty, so the assessment asks two things: is there a rigorous link between the two problems, and how strong is it.
The link. Tao's 2011 post (terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/) isolates the "weak Collatz conjecture" (no cycle other than 1, 4, 2), proves its equivalence with the Böhm–Sontacchi divisibility formulation (Proposition 4), and then proves Proposition 6: if the weak conjecture holds, then 2^n − 3^m, for 2^n > 3^m, is bounded below by a quantity growing with n. The mechanism is that the set of admissible sums of products of powers of 2 and 3 contains parallelepipeds of dimension comparable to n whose generators are coprime to the modulus 2^n − 3^m, and such a parallelepiped in a cyclic group of order smaller than its dimension would cover every residue, including the forbidden one. Proposition 7, sketched via Riesz products and Bohr-set concentration, improves the bound. This is the subclaim that the absence of nontrivial Collatz cycles implies a nontrivial lower bound on |2^n − 3^m|; the elementary version is fully argued and has stood unchallenged since 2011, though it is a blog proof rather than a refereed one.
The state of the art on the other side. Effective lower bounds for |2^n − 3^m| that grow with n come from the theory of linear forms in two logarithms (Gelfond, then Baker; Baker's version gives 2^n divided by a fixed power of n). Even the ineffective finiteness of solutions to 2^n − 3^m = k for each k (Pillai) rests on the Thue–Siegel theorem, which belongs to the same Diophantine-approximation tradition. No elementary argument for a bound tending to infinity is known; Tao states he knows of none, and his 2020 lecture slides (terrytao.wordpress.com/wp-content/uploads/2020/04/collatz-1.pdf) repeat that all known unconditional proofs require some variant of Baker's theorem. This is the subclaim that every known nontrivial lower bound relies on transcendence theory, which the comparison needs: without it, a Collatz proof would not have to reproduce transcendence-theoretic work. It is a claim about the literature and would fall to a single elementary proof, but none was found.
How strong the link is. Tao is explicit that Proposition 6 "is very weak when compared against the unconditional bound" and that Proposition 7 is "still well short of the transcendence theory bound". So what Collatz implies is a nontrivial separation, not Baker's theorem, and the subclaim that the implied bound is far weaker than Baker's weighs against the literal reading of "at least as difficult as". Tao's own careful conclusion is that any proof "must either use existing results in transcendence theory, or else must contribute a new method to give non-trivial results in transcendence theory"; the 2019 phrasing that motivates this claim adds "basically" as a hedge. Maxwell Siegel's expository paper (arxiv.org/pdf/2412.02902) states that a proof of the weak conjecture "would necessarily yield a proof of Baker's Theorem far simpler than any currently known method", which the cited argument does not deliver.
Corroboration. The known exclusions of structured cycles (Steiner's 1-cycles, Simons and de Weger's m-cycles) all use Baker-type bounds, the subclaim that all known proofs excluding m-cycles rely on Baker's theorem; this supports the view that the cycle problem is genuinely entangled with transcendence theory.
Instances and weighing. Three instances, all affirming, but effectively one voice: Tao's 2011 argument, his 2019 restatement, and a 2024 repetition that strengthens it. No source denies the claim, and the search for a dissenting or elementary route found none. The verdict is "supported" rather than "verified" because the comparison is informal: its rigorous core (a Collatz proof yields nontrivial transcendence-type results) is established, while the literal comparative with Baker's theorem outruns the proved implication. What would move the verdict: an elementary proof of a growing lower bound for |2^n − 3^m| would undercut the claim; a proof that the Collatz conjecture implies Baker-strength separation would make the literal reading verified; a proof of Collatz by methods that yield no transcendence-theoretic result would contradict it outright. No credence is given because the claim is a comparative judgment rather than a proposition with a single truth value.
Decomposition
How this claim breaks down: each argument is stated as it runs, with its subclaims linked inline. ↗︎ opens a subclaim; the map shows how they fit together.
Because the absence of nontrivial Collatz cycles already implies a nontrivial lower bound on |2^n − 3^m|, any proof of the Collatz conjecture would deliver such a bound as a corollary; and since every known nontrivial lower bound of this kind rests on transcendence theory, such a proof would have to either invoke Baker-type results or supply a new method capable of replacing them. That all known exclusions of structured Collatz cycles already run through Baker's theorem corroborates that the entanglement is real rather than formal.
The inference goes through for the qualified conclusion that a proof of Collatz must use or replace transcendence theory: it needs both the conditional lower bound on the gap between powers of 2 and 3, which Tao proved by an elementary argument, and the absence of any known non-transcendental route to such bounds, a claim about the literature that is well attested but would fall to a single elementary proof. It does not deliver the literal conclusion that a Collatz proof is at least as hard as Baker's theorem, because the bound it produces is much weaker than Baker's; the corroborating premise about m-cycle exclusions adds plausibility without closing that gap.
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.
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.
Provenance
Where this claim has been said, linked to its canonical form.
The support is one voice. Terence Tao argued the comparison in a 2011 blog post, proving that the absence of nontrivial Collatz cycles forces a nontrivial gap between powers of 2 and 3 and observing that such gaps are otherwise known only through transcendence theory; his 2019 paper announcement restates that conclusion in a compressed form, with "basically" as the hedge. A later expository paper repeats the point with the hedge dropped, asserting that a proof of the weak conjecture would yield Baker's theorem itself, which the cited argument does not show. A reader should open the 2011 post, where Tao himself notes that the conditional bounds fall well short of Baker's.
Thus, this result strongly suggests that any proof of the Collatz conjecture must either use existing results in transcendence theory, or else must contribute a new method to give non-trivial results in transcendence theory.
The originating argument: after showing that the no-nontrivial-cycles consequence of Collatz implies a nontrivial lower bound on 2^n − 3^m (Propositions 6 and 7), Tao draws the conclusion about what any proof of Collatz must involve. This is the careful form of the later remark that Collatz is "basically at least as difficult as Baker's theorem"; it stops short of the literal comparison, hence the reduced confidence.
The source's own evidence bears what it asserts. This is the origin of the comparison. The post proves the weaker conditional bound in full and sketches a stronger one, and it is careful to say that both fall well short of Baker's theorem; the claim it actually argues for is that any proof of Collatz must use or replace transcendence theory, not that it would reprove Baker's theorem. Worth reading closely: It is the only source that argues the comparison rather than stating it; Propositions 6 and 7 and the surrounding remarks fix exactly how strong the Collatz-to-transcendence link is.
Any proof of the Weak Collatz Conjecture will necessarily entail a significant advancement in transcendental number theory. A proof of the Weak Collatz Conjecture would necessarily yield a proof of Baker's Theorem far simpler than any currently known method [147].
A section titled "Baker, Catalan, and Collatz" in a long expository thesis-style paper, citing Tao's 2011 blog post; it states the comparison more strongly than Tao does, asserting that a proof of the weak conjecture would yield a proof of Baker's theorem itself.
Only the passage itself was read, not the surrounding document. The passage states the comparison more strongly than its cited source does, asserting that a proof of the weak conjecture would yield a proof of Baker's theorem, whereas the cited argument yields a much weaker bound. Worth reading closely: To see whether the author offers any argument of his own for the stronger statement or simply restates Tao's remark with the qualification dropped.
it is basically at least as difficult as Baker’s theorem, all known proofs of which are quite difficult
Establishing the Collatz conjecture for all N remains out of reach of current techniques
Asserted without evidence of the source's own. The remark is a one-line aside in a paper announcement; the argument behind it lives in the author's earlier post on Collatz and powers of 2 and 3, which this post cites rather than repeats. The hedge "basically" is the author's own. The quoted passage was not found in the stored copy of this source.
How these sources relate
- https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-orbits-attain-almost-bounded-values/ restates https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/, stating it more strongly than that document supports. The 2019 remark restates the 2011 argument's conclusion in compressed form. The 2011 post concludes only that a proof must use or replace transcendence theory and says its conditional bounds fall well short of Baker's; "at least as difficult as Baker's theorem" is a stronger gloss, softened by "basically".
- (p,q)-adic Analysis and the Collatz Conjecture draws its statement from https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/, stating it more strongly than that document supports. The statement is attributed to a reference, and the only published argument of this shape is Tao's 2011 post, whose conditional bounds are much weaker than Baker's theorem; stating that a Collatz proof would yield Baker's theorem itself drops that qualification. The reference number was not checked against the bibliography, so the target identification rests on the content.
- https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/ and https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-orbits-attain-almost-bounded-values/ share an author. Both are posts on Terence Tao's blog "What's new", bylined "by Terence Tao" (2019 and 2011).
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Created by extractor · Sep 13, 2026. Every judgment on this page is accompanied by a reasoning trace.