FormalConjectures/ErdosProblems/405.lean
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3 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 405
References:
- erdosproblems.com/405
- [ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathematique (1980)
- [BrEr91] Brindza, B. and Erdős, P., On some {D}iophantine problems involving powers and factorials. J. Austral. Math. Soc. Ser. A (1991), 1--7.
- [YuLi96] Yu, Kunrui and Liu, Dehua, A complete resolution of a problem of {E}rdős and {G}raham. Rocky Mountain J. Math. (1996), 1235--1244.
Imports (1)
- theorem
Erdos405.erdos_405Let $p$ be an odd prime sorry line 48 sourceSource documentation
Let $p$ be an odd prime. Is it true that the equation $(p-1)!+a^{p-1}=p^k$ has only finitely many solutions?
Originally proposed by Erdős and Graham [ErGr80]. Brindza and Erdős [BrEr91] proved that there are finitely many such solutions.
depends on sorry what this means
Statement uses:
Set.Finite,Prod,Nat,Set.ofPred,Erdos405.erdos_405.match_1,And,Nat.Prime,Odd,Eq,HAdd.hAdd,Nat.factorial,HSub.hSub,OfNat.ofNat,HPow.hPowProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous20 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_405 : Set.Finite { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧ Odd p ∧ … - theorem
Erdos405.erdos_405.variants.nonodd_pErdős and Graham [ErGr80] ask this allowing $p=2$, but this is presumably an oversight, since clearly there are infinitely many solutions when $p=2$ line 60 sourceSource documentation
Erdős and Graham [ErGr80] ask this allowing $p=2$, but this is presumably an oversight, since clearly there are infinitely many solutions when $p=2$.
Observe that this creates $1! + a = 2^k$. For all $k$, fix $a = 2^k - 1$.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
Set.Infinite,Prod,Nat,Set.ofPred,Erdos405.erdos_405.match_1,And,Nat.Prime,Eq,HAdd.hAdd,Nat.factorial,HSub.hSub,OfNat.ofNat,HPow.hPowProof uses:
Set.infinite_of_injective_forall_mem,Prod.mk,And.left,True,And.right,Eq.mp,Eq.trans,Prod.mk.injEq,congrArg,eq_self,id,Membership.mem,Set,And.intro,Nat.prime_two,Eq.mpr,pow_one,LT.lt,pow_pos,Mathlib.Meta.Positivity.pos_of_isNatand 6 more35 instances, coercions and auxiliary constants left out.
Tactic trace: 14 steps
Open to load the checked before-and-after goal states.
theorem erdos_405.variants.nonodd_p : Set.Infinite { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧ … - theorem
Erdos405.erdos_405.variants.yu_liuYu and Liu [YuLi96] showed that the only solutions to $(p-1)! + a^(p-1) = p^k$ for an odd prime p are: $$ 2! + 1^2 = 3\\ 2! + 5^2 = 3^3\\ 4! + 1^4 = 5^2 $$ sorry line 94 sourceSource documentation
Yu and Liu [YuLi96] showed that the only solutions to $(p-1)! + a^(p-1) = p^k$ for an odd prime p are: $$ 2! + 1^2 = 3\\ 2! + 5^2 = 3^3\\ 4! + 1^4 = 5^2 $$
depends on sorry what this means
Statement uses:
Eq,Set,Prod,Nat,Set.ofPred,Erdos405.erdos_405.match_1,And,Nat.Prime,Odd,HAdd.hAdd,Nat.factorial,HSub.hSub,OfNat.ofNat,HPow.hPow,Insert.insert,Prod.mk,Singleton.singletonProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous24 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_405.variants.yu_liu : { x : ℕ × ℕ × ℕ | let (a, k, p) := x; p.Prime ∧ Odd p ∧ …