@sat0shinakomoto: The infinitude of primes using Fibonacci and/or Benford's Law. by Satoshi #physics #tpatrickmurray #mathematics #satoshinakamoto #bitcoin The Fibonacci-Benford Proof of the Infinitude of Primes I. Euclid's Classical Proof (c. 300 BCE) First, the standard. Assume finitely many primes: p₁, p₂, ..., pₙ. Form N = p₁·p₂·...·pₙ + 1. N is either prime (contradiction) or has a prime factor not in the list (contradiction). Therefore infinitely many primes. QED. Elegant. But it reveals nothing about why primes are infinite. It's a logical trick, not a structural explanation. II. The Fibonacci-Golden Ratio Proof Theorem: The Fibonacci sequence, scaled by the golden ratio φ, requires infinitely many primes for its multiplicative generation. Proof: The Fibonacci numbers satisfy Binet's formula: F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}} where φ = (1+√5)/2. For large n, F_n ≈ φⁿ/√5. Now consider the set of all prime-indexed Fibonacci numbers: S = {F_p : p prime}. Lemma 1: If p and q are distinct primes, then F_p and F_q are relatively prime. Proof: It's a known theorem that gcd(F_m, F_n) = F_{gcd(m,n)}. Since gcd(p,q) = 1 for distinct primes, gcd(F_p, F_q) = F_1 = 1. ∎ Lemma 2: Distinct prime-indexed Fibonacci numbers contribute distinct prime factors not shared with any smaller prime-indexed Fibonacci number. Proof: From Lemma 1, F_p shares no common factors with any F_q for q ≠ p. Since F_p > 1 for all p > 2, each F_p introduces at least one new prime factor not present in any F_q for q < p. ∎ Theorem: There are infinitely many primes. Proof by contradiction: Assume finitely many primes: p₁, p₂, ..., pₙ. Then the set of prime-indexed Fibonacci numbers {F_{p₁}, F_{p₂}, ..., F_{pₙ}} is finite. But each such F_{pᵢ} requires at least one distinct prime factor (Lemma 2). Therefore at least n distinct prime factors exist. But we can always generate F_{pₙ₊₁} for any new prime index pₙ₊₁, which requires a new prime factor. Since the indexing primes are inexhaustible (contradiction of finiteness), the prime factors are inexhaustible. ∎ Wait—that's circular. It assumes infinitely many indexing primes to prove infinitely many primes. Let me fix this. III. The Corrected Fibonacci Proof (Eliminating Circularity) Theorem: There are infinitely many primes. Proof: Assume finitely many primes: p₁, ..., pₖ. Consider the Fibonacci numbers F_n for n = 1, 2, 3, ... By the gcd property: gcd(F_m, F_n) = F_{gcd(m,n)}. Take any two Fibonacci numbers with coprime indices m and n (gcd(m,n)=1). Then gcd(F_m, F_n) = F_1 = 1. So F_m and F_n are coprime, meaning they share no prime factors. Now take the infinite set of Fibonacci numbers with indices that are powers of 2: n = 2¹, 2², 2³, 2⁴, ... gcd(2ⁱ, 2ʲ) = 2^{min(i,j)} ≠ 1, so this doesn't guarantee coprimality. Let me use odd numbers. Take n = 1, 2, 3, 5, 7, 11, ... — but that assumes primes. Let me use coprime indices differently. Take n = 1, 2, 3, 4, 5, 6, 7, 8, ... Consider the infinite subset where indices are pairwise coprime. For example, let n run through the sequence: 2, 3, 5, 7, 11, 13, 17, 19, 23, ... — that's primes again. I'm circling. Let me try a different construction. IV. Non-Circular Fibonacci Construction Take the sequence of Fibonacci numbers at indices n = 2^k for k = 1, 2, 3, 4, ... Consider: gcd(2ⁱ, 2ʲ) = 2^{min(i,j)}. So gcd(F_{2ⁱ}, F_{2ʲ}) = F_{2^{min(i,j)}}. This doesn't guarantee coprimality. But F_{2^k} grows super-exponentially and each introduces new prime factors because the set of prime divisors of F_{2^k} is strictly larger than the set of prime divisors of F_{2^{k-1}} in infinitely many cases. This is true but slow to prove. Let me use the correct non-circular argument. V. The Actual Elegant Pr

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Wednesday 29 April 2026 08:32:31 GMT