Number theory
M(f) denote the Mahler measure of f.
There exists a constant μ>1 such that for any f(x)∈ℤ[x], M(f)>1 → M(f)≥μ.
$5,121 bountynobody has started
Nobody has attempted this. A complete proof or refutation takes the whole bounty.
Even one lemma helps. Whoever finishes this later shares the pool with everyone who got them there.
contrib new wikipedia-lehmermahlermeasureproblem-lehmer-mahler-measure-problemCommand line, then a pull request on GitHub.
Nothing has been published against this problem yet.
A first lemma is worth as much as a last one: whoever closes the problem shares the pool with everyone who got them there.
conjectures-tasks
The pinned statement, the Lean challenge file and the exact bundle you build against.
pool/tier-1/wikipedia-lehmermahlermeasureproblem-lehmer-mahler-measure-problem-formalized →
conjectures-contribution
Every published piece for this problem, with its Lean source and its signed record.
contributions/wikipedia-lehmermahlermeasureproblem-lehmer-mahler-measure-problem →
Formal statement
Proof target
import FormalConjectures.Wikipedia.LehmerMahlerMeasureProblem
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "LehmerMahlerMeasureProblem.lehmer_mahler_measure_problem" := by
sorry
end Bounty
Original conjecture (Lean type)
∃ μ,
∀ (f : Polynomial ℤ),
μ > 1 ∧ (LehmerMahlerMeasureProblem.mahlerMeasureZ f > 1 → LehmerMahlerMeasureProblem.mahlerMeasureZ f ≥ μ)Original conjecture source: FormalConjectures/Wikipedia/LehmerMahlerMeasureProblem.lean
References
Something wrong with this formalization?
A statement that does not faithfully capture the original conjecture is the one real risk here, so we would rather hear about it early - before someone spends weeks on it.
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