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Formalizations.NumberTheory.Fibonacci.SumDivisibility

Infinitely many odd k divide F_1 + ⋯ + F_k (Flynn-Connolly 2025) #

For an odd prime p with p ≡ 2 (mod 3) and p ≡ ±2 (mod 5), and n = 2p or n = 4p, the Fibonacci number F_n is odd and F_{F_n + 2} ≡ 1 (mod F_n), so F_n divides F_1 + ⋯ + F_{F_n} = F_{F_n + 2} − 1; by Dirichlet's theorem there are infinitely many odd k with k ∣ F_1 + ⋯ + F_k, the conjecture of Fatehizadeh and Yaqubi (O. Flynn-Connolly, INTEGERS 25 (2025) A87). The statements are those of atlas-lean's FibonacciSumDivisibilityWanted. Proof: for even n the Pisano period of F_n divides 2n (Cassini); for such p, p ∣ F_{p+1} (from F_p ≡ (5/p), proved earlier in this library), so the period modulo 4p and 8p divides 2(p + 1).

Main results #

References #

F_{k+T} ≡ F_k (mod m) for all k.

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Instances For
    theorem FibonacciSumDivisibility.per_of_return {m : ℕ} (T : ℕ) (h0 : ↑(Nat.fib T) = 0) (h1 : ↑(Nat.fib (T + 1)) = 1) :
    Per m T
    theorem FibonacciSumDivisibility.per_mul {m T : ℕ} (h : Per m T) (j : ℕ) :
    Per m (j * T)
    theorem FibonacciSumDivisibility.per_crt {a b T : ℕ} (hc : a.Coprime b) (h1 : Per a T) (h2 : Per b T) :
    Per (a * b) T
    theorem FibonacciSumDivisibility.period_fib_even (n : ℕ) (hn : Even n) (_hn1 : 1 < n) (k : ℕ) :

    Proposition 2 (Flynn-Connolly): for even n > 1, F_{k+2n} ≡ F_k (mod F_n).

    theorem FibonacciSumDivisibility.two_fib_succ (p : ℕ) [Fact (Nat.Prime p)] (hp2 : p ≠ 2) :
    2 * ↑(Nat.fib (p + 1)) = 5 ^ (p / 2) + 1

    The (0,0) entry of the Frobenius identity for the Fibonacci matrix (copy of the method of FibLegendre.fib_prime_eq_legendreSym, whose (0,1) entry gives F_p): 2 F_{p+1} = 5^{(p-1)/2} + 1.

    theorem FibonacciSumDivisibility.legendreSym_five (p : ℕ) [Fact (Nat.Prime p)] (hp2 : p ≠ 2) (h5 : p % 5 = 2 ∨ p % 5 = 3) :
    legendreSym p 5 = -1

    legendreSym p 5 = -1 for a prime p ≡ ±2 (mod 5).

    theorem FibonacciSumDivisibility.fib_succ_prime_dvd (p : ℕ) (hp : Nat.Prime p) (h5 : p % 5 = 2 ∨ p % 5 = 3) :
    p ∣ Nat.fib (p + 1) ∧ ↑(Nat.fib p) = -1

    Wall (case (5/p) = -1): for a prime p ≡ ±2 (mod 5), F_p ≡ -1 and p ∣ F_{p+1}.

    theorem FibonacciSumDivisibility.per_p (p : ℕ) (hp : Nat.Prime p) (h5 : p % 5 = 2 ∨ p % 5 = 3) :
    Per p (2 * p + 2)

    The period 2p + 2 = 2(p+1) modulo p.

    theorem FibonacciSumDivisibility.per_4p {p : ℕ} (hp : Nat.Prime p) (hpodd : p % 2 = 1) (hp3 : p % 3 = 2) (hp5 : p % 5 = 2 ∨ p % 5 = 3) :
    Per (4 * p) (2 * p + 2)
    theorem FibonacciSumDivisibility.per_8p {p : ℕ} (hp : Nat.Prime p) (hpodd : p % 2 = 1) (hp3 : p % 3 = 2) (hp5 : p % 5 = 2 ∨ p % 5 = 3) :
    Per (8 * p) (2 * p + 2)
    theorem FibonacciSumDivisibility.fib_two_mul_mod {p : ℕ} (hp : Nat.Prime p) (hpodd : p % 2 = 1) (hp3 : p % 3 = 2) (hp5 : p % 5 = 2 ∨ p % 5 = 3) :
    4 * p ∣ Nat.fib (2 * p) + 1

    F_{2p} ≡ -1 (mod 4p).

    theorem FibonacciSumDivisibility.fib_four_mul_mod {p : ℕ} (hp : Nat.Prime p) (hpodd : p % 2 = 1) (hp3 : p % 3 = 2) (hp5 : p % 5 = 2 ∨ p % 5 = 3) :
    8 * p ∣ Nat.fib (4 * p) + 3

    F_{4p} ≡ -3 (mod 8p).

    F_n is odd when 3 ∤ n.

    theorem FibonacciSumDivisibility.fibfib_mod (n : ℕ) (hn : Even n) (hn4 : 4 ≤ n) (h : 2 * n ∣ Nat.fib n + 1 ∨ 2 * n ∣ Nat.fib n + 3) :

    Core step: if F_n has period 2n and 2n ∣ F_n + 1 or 2n ∣ F_n + 3 (n ≥ 4 even) then F_{F_n+2} ≡ 1 (mod F_n).

    theorem FibonacciSumDivisibility.theorem_two (p n : ℕ) (hp : Nat.Prime p) (hpodd : p % 2 = 1) (hp3 : p % 3 = 2) (hp5 : p % 5 = 2 ∨ p % 5 = 3) (hn : n = 2 * p ∨ n = 4 * p) :
    Nat.fib n % 2 = 1 ∧ Nat.fib (Nat.fib n + 2) % Nat.fib n = 1

    Theorem 2 (Flynn-Connolly) for a prime p.

    theorem FibonacciSumDivisibility.flynn_connolly :
    (∀ (p n : ℕ), Nat.Prime p → p % 2 = 1 → p % 3 = 2 → p % 5 = 2 ∨ p % 5 = 3 → n = 2 * p ∨ n = 4 * p → Nat.fib n % 2 = 1 ∧ Nat.fib (Nat.fib n + 2) % Nat.fib n = 1) ∧ {k : ℕ | Odd k ∧ k ∣ ∑ i ∈ Finset.Icc 1 k, Nat.fib i}.Infinite

    Flynn-Connolly (2025). Theorem 2 and Corollary 2.