Prime divisors of have the form
ProvedAlfutovaUstinov.problem_4_122elementary-number-theorymersenne-numbersmultiplicative-ordernumber-theory
This is Problem 4.122 of N. B. Alfutova and A. V. Ustinov, Algebra and Number Theory (MCCME, 2002), Chapter 4, §4 “Theorems of Fermat and Euler”.
Theorem. Let be a prime number. Then every prime divisor of the Mersenne number has the form
for some natural number .
This classical fact about Mersenne numbers (going back to Fermat and Euler) drastically restricts the possible prime factors of and is used when searching for Mersenne primes.
Formalization Note Here and are natural numbers with Nat.Prime; the subtraction is natural-number subtraction, which is exact since .
Preamble
import Mathlib
Formal statement
namespace AlfutovaUstinov
theorem problem_4_122 (p : ℕ) (hp : p.Prime) (hp2 : 2 < p) (q : ℕ) (hq : q.Prime)
(hqd : q ∣ 2 ^ p - 1) : ∃ k : ℕ, q = 2 * k * p + 1 := by sorry
end AlfutovaUstinovSource
N. B. Alfutova, A. V. Ustinov, «Алгебра и теория чисел. Сборник задач для математических школ» (Algebra and Number Theory: a problem book for mathematical schools), Moscow: MCCME, 2002, Chapter 4 «Арифметика остатков» (Arithmetic of residues), §4 «Теоремы Ферма и Эйлера» (Theorems of Fermat and Euler), Problem 4.122. Problem text and answer as catalogued on problems.ru, problem 60748: https://problems.ru/view_problem_details_new.php?id=60748