Büchi's problem

References:

• Wikipedia

def Buchi.IsBuchi (M : ℕ) : Prop

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.

Equations

• Buchi.IsBuchi M = ∀ (x a : ℤ), (∀ n ∈ Finset.range M, IsSquare ((x + ↑n) ^ 2 + a)) → a = 0

theorem Buchi.buchi_problem : True ↔ ∃ (M : ℕ), 1 ≤ M ∧ IsBuchi M

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 $a = 0$.

theorem Buchi.buchi_problem_M5 : IsBuchi 5

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.

theorem Buchi.buchi_false_M0 : ¬IsBuchi 0

The case $M = 0$ fails: the hypothesis is vacuous, so $a = 0$ cannot be concluded.

theorem Buchi.buchi_false_M1 : ¬IsBuchi 1

The case $M = 1$ fails: $0^2 + 4 = 4 = 2^2$ is a square but $a = 4 \neq 0$.

theorem Buchi.buchi_false_M2 : ¬IsBuchi 2

The case $M = 2$ fails: $(7+n)^2 - 48$ is a perfect square for $n = 0, 1$ ($1^2$ and $4^2$), but $a = -48 \neq 0$.

theorem Buchi.buchi_false_M3 : ¬IsBuchi 3

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 $a = -576 \neq 0$.

theorem Buchi.buchi_false_M4 : ¬IsBuchi 4

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).