Buchi.lean`

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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

8 declarations, 2 contain sorry, 0 declared as axioms

Büchi's problem

References:

Wikipedia

Imports (1)

FormalConjecturesUtil

Imported by 1 file

FormalConjectures/Subsets/FC100OpenSet1.lean

def Buchi.IsBuchi IsBuchi M asserts that whenever M consecutive values (x + n)² + a (for n = 0, …, M - 1) are all perfect squares, then a must be 0 line 33 source
Source documentation
IsBuchi M asserts that whenever M consecutive values (x + n)² + a (for n = 0, …, M - 1) are all perfect squares, then a must be 0.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Nat
Proof uses: Int, Membership.mem, Finset, Finset.range, IsSquare, HAdd.hAdd, HPow.hPow, OfNat.ofNat, Eq
13 instances, coercions and auxiliary constants left out.
```
def IsBuchi (M : ℕ) : Prop :=
…
```
theorem Buchi.buchi_problem Büchi's problem There exists a positive integer $M$ such that, for all integers $x$ and $a$, if $(x+n)^2 + a$ is a square for $M$ consecutive values of $n$, then $a = 0$ sorry line 42 source
Source documentation
Büchi's problem There exists a positive integer $M$ such that, for all integers $x$ and $a$, if $(x+n)^2 + a$ is a square for $M$ consecutive values of $n$, then $a = 0$.

depends on sorry what this means
Statement uses: Iff, True, Exists, Nat, And, LE.le, OfNat.ofNat, Buchi.IsBuchi
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
4 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_problem :
    answer(sorry) ↔ ∃ M : ℕ, 1 ≤ M ∧ IsBuchi M := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_problem_M5 Büchi's problem (first open case, $M = 5$): For all integers $x$ and $a$, if $(x+n)^2 + a$ is a perfect square for $n = 0, 1, 2, 3, 4$, then $a = 0$ sorry line 55 source
Source documentation
Büchi's problem (first open case, $M = 5$): For all integers $x$ and $a$, if $(x+n)^2 + a$ is a perfect square for $n = 0, 1, 2, 3, 4$, then $a = 0$.
Non-trivial sequences of length 3 and 4 are known to exist, so $M = 5$ is the first open case.

depends on sorry what this means
Statement uses: Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
2 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_problem_M5 : IsBuchi 5 := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_false_M0 The case $M = 0$ fails: the hypothesis is vacuous, so $a = 0$ cannot be concluded line 60 source
Source documentation
The case $M = 0$ fails: the hypothesis is vacuous, so $a = 0$ cannot be concluded.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: Eq.mpr, Exists, Int, And, Membership.mem, Finset, EmptyCollection.emptyCollection, IsSquare, HAdd.hAdd, HPow.hPow, Eq, id, Eq.trans, congrArg, funext, Exists.intro, of_eq_true, True, congr, forall_congr and 15 more
26 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_false_M0 : ¬ IsBuchi 0 := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_false_M1 The case $M = 1$ fails: $0^2 + 4 = 4 = 2^2$ is a square but $a = 4 \neq 0$ line 66 source
Source documentation
The case $M = 1$ fails: $0^2 + 4 = 4 = 2^2$ is a square but $a = 4 \neq 0$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: Eq.mpr, Exists, Int, Membership.mem, Finset, Finset.range, IsSquare, HAdd.hAdd, HPow.hPow, Eq, id, Eq.trans, congrArg, funext, Exists.intro, HMul.hMul, Mathlib.Meta.NormNum.isNat_eq_true, Mathlib.Meta.NormNum.isNat_add, Eq.refl, Mathlib.Meta.NormNum.isNat_pow and 9 more
29 instances, coercions and auxiliary constants left out.

Tactic trace: 8 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_false_M1 : ¬ IsBuchi 1 := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_false_M2 The case $M = 2$ fails: $(7+n)^2 - 48$ is a perfect square for $n = 0, 1$ ($1^2$ and $4^2$), but $a = -48 \neq 0$ line 76 source
Source documentation
The case $M = 2$ fails: $(7+n)^2 - 48$ is a perfect square for $n = 0, 1$ ($1^2$ and $4^2$), but $a = -48 \neq 0$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: Eq.mpr, Exists, Int, Membership.mem, Finset, Finset.range, IsSquare, HAdd.hAdd, HPow.hPow, Eq, id, Eq.trans, congrArg, funext, Exists.intro, Neg.neg, dite, LE.le, HMul.hMul, Mathlib.Meta.NormNum.isNat_eq_true and 27 more
54 instances, coercions and auxiliary constants left out.

Tactic trace: 12 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_false_M2 : ¬ IsBuchi 2 := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_false_M3 The case $M = 3$ fails: $(24+n)^2 - 576$ is a perfect square for $n = 0, 1, 2$ ($0^2$, $7^2$, and $10^2$), but $a = -576 \neq 0$ line 88 source
Source documentation
The case $M = 3$ fails: $(24+n)^2 - 576$ is a perfect square for $n = 0, 1, 2$ ($0^2$, $7^2$, and $10^2$), but $a = -576 \neq 0$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: Eq.mpr, Exists, Int, Membership.mem, Finset, Finset.range, IsSquare, HAdd.hAdd, HPow.hPow, Eq, id, Eq.trans, congrArg, funext, Exists.intro, Neg.neg, dite, LE.le, LT.lt, HMul.hMul and 27 more
43 instances, coercions and auxiliary constants left out.

Tactic trace: 15 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_false_M3 : ¬ IsBuchi 3 := by
…
```
Mentions: Buchi.IsBuchi
theorem Buchi.buchi_false_M4 The case $M = 4$ fails: $(246+n)^2 - 60480$ is a perfect square for $n = 0, 1, 2, 3$ ($6^2$, $23^2$, $32^2$, and $39^2$), but $a = -60480 \neq 0$ line 102 source
Source documentation
The case $M = 4$ fails: $(246+n)^2 - 60480$ is a perfect square for $n = 0, 1, 2, 3$ ($6^2$, $23^2$, $32^2$, and $39^2$), but $a = -60480 \neq 0$. The sequence $(6, 23, 32, 39)$ is a nontrivial Büchi sequence of length 4 (Hensley 1983).

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Not, Buchi.IsBuchi, OfNat.ofNat, Nat
Proof uses: Eq.mpr, Exists, Int, Membership.mem, Finset, Finset.range, IsSquare, HAdd.hAdd, HPow.hPow, Eq, id, Eq.trans, congrArg, funext, Exists.intro, Neg.neg, dite, LE.le, LT.lt, HMul.hMul and 27 more
43 instances, coercions and auxiliary constants left out.

Tactic trace: 18 steps
Open to load the checked before-and-after goal states.
```
theorem buchi_false_M4 : ¬ IsBuchi 4 := by
…
```
Mentions: Buchi.IsBuchi