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module
public import FormalConjecturesUtilErdős Problem 477
References:
Sek59 Milan Sekanina, Замечания к фактoризации беcкoнечнoй цикличеcкoй группы, Czechoslovak Mathematical Journal, Vol. 9 (1959), No. 4, 485–495
The resolution is recorded at erdosproblems.com/477, with a proof exposition by T. F. Bloom of a construction found independently by several provers.
@[expose] public sectionopen Polynomial Setnamespace Erdos477Is there a polynomial $f:\mathbb{Z}\to \mathbb{Z}$ of degree at least $2$ and a set $A\subset \mathbb{Z}$ such that for any $n\in \mathbb{Z}$ there is exactly one $a\in A$ and $b\in { f(k) : k\in\mathbb{Z}}$ such that $n=a+b$?
The answer is yes, contrary to the expectation of Erdős and Graham: such an A exists whenever
$f(n) = n^d$ for even $d \ge 6$.
@[category research solved, AMS 12]
theorem erdos_477 : answer(True) ↔
∃ f : ℤ[X], 2 ≤ f.degree ∧ ∃ A : Set ℤ,
∀ z, ∃! ab ∈ A ×ˢ (Set.range f.eval), z = ab.1 + ab.2 := ⊢ True ↔ ∃ f, 2 ≤ f.degree ∧ ∃ A, ∀ (z : ℤ), ∃! ab, (ab ∈ A ×ˢ range fun x ↦ eval x f) ∧ z = ab.1 + ab.2
All goals completed! 🐙There is no such $A$ for the polynomial $f(x) = X^2$.
This is shown in [Sek59].
@[category research solved, AMS 12]
theorem erdos_477.variants.S_sq :
letI f := X ^ 2
∀ A : Set ℤ, ∃ z, ¬ ∃! a ∈ A ×ˢ (Set.range f.eval), z = a.1 + a.2 := ⊢ ∀ (A : Set ℤ), ∃ z, ¬∃! a, (a ∈ A ×ˢ range fun x ↦ eval x (X ^ 2)) ∧ z = a.1 + a.2
All goals completed! 🐙There is no such $A$ for any polynomial $f(x) = aX^2 + bX + c$, if $a | b$ with $a \ne 0$ and $b \ne 0$. This was found be AlphaProof for the specific instance $X^2 - X + 1$ and then generalised.
@[category research solved, AMS 12]
theorem erdos_477.variants.degree_two_dvd_condition_b_ne_zero {a b c : ℤ} (ha : a ≠ 0) (hb : b ≠ 0)
(hab : a ∣ b) :
let f := a • X ^ 2 + b • X + C c
∀ A : Set ℤ, ∃ z, ¬ ∃! a ∈ A ×ˢ (Set.range f.eval), z = a.1 + a.2 := a:ℤb:ℤc:ℤha:a ≠ 0hb:b ≠ 0hab:a ∣ b⊢ let f := a • X ^ 2 + b • X + C c;
∀ (A : Set ℤ), ∃ z, ¬∃! a, (a ∈ A ×ˢ range fun x ↦ eval x f) ∧ z = a.1 + a.2
All goals completed! 🐙Probably there is no such $A$ for the polynomial $X^3$.
@[category research open, AMS 12]
theorem erdos_477.variants.X_pow_three :
letI f := X ^ 3
∀ A : Set ℤ, ∃ z, ¬ ∃! a ∈ A ×ˢ (Set.range f.eval), z = a.1 + a.2 := ⊢ ∀ (A : Set ℤ), ∃ z, ¬∃! a, (a ∈ A ×ˢ range fun x ↦ eval x (X ^ 3)) ∧ z = a.1 + a.2
All goals completed! 🐙Sekanina [Sek59] asked whether there is no such $A$ for $X^k$, for every $k \ge 2$. This is false: a complement exists for every even $k \ge 6$.
@[category research solved, AMS 12]
theorem erdos_477.variants.monomial : answer(False) ↔
∀ (k : ℕ), 2 ≤ k →
letI f := X ^ k
∀ A : Set ℤ, ∃ z, ¬ ∃! a ∈ A ×ˢ (Set.range f.eval), z = a.1 + a.2 := ⊢ False ↔ ∀ (k : ℕ), 2 ≤ k → ∀ (A : Set ℤ), ∃ z, ¬∃! a, (a ∈ A ×ˢ range fun x ↦ eval x (X ^ k)) ∧ z = a.1 + a.2
All goals completed! 🐙end Erdos477