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gcd_sum_diff.lean
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79 lines (60 loc) · 2.04 KB
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import Mathlib
--
-- Outline of proof
--
-- m is odd
-- n is even
-- gcd(m, n) = 1
-- Let d = gcd(m + n, m - n)
-- d divides (m + n) + (m - n) = 2*m
-- d divides (m + n) - (m - n) = 2*n
-- gcd(2*m, 2*n) = 2*gcd(m, n) = 2
-- So d is 1 or 2
-- But m + n and m - n are both odd
-- Therefore d is 1
theorem gcd_sum_diff (m n : ℤ)
(h_m_odd : m % 2 = 1)
(h_n_even : n % 2 = 0)
(h_coprime : m.gcd n = 1) :
(m + n).gcd (m - n) = 1 := by
have h_sum_odd : (m + n) % 2 = 1 := by
rw [Int.add_emod, h_m_odd, h_n_even]
norm_num
have h_diff_odd : (m - n) % 2 = 1 := by
rw [Int.sub_emod, h_m_odd, h_n_even]
norm_num
let d := (m + n).gcd (m - n)
have hd_dvd_sum : (d : ℤ) ∣ (m + n) := Int.gcd_dvd_left _ _
have hd_dvd_diff : (d : ℤ) ∣ (m - n) := Int.gcd_dvd_right _ _
-- d divides (m + n) + (m - n) = 2*m
have h_d_dvd_2m : (d : ℤ) ∣ 2 * m := by
have h1 : (d : ℤ) ∣ (m + n) + (m - n) := dvd_add hd_dvd_sum hd_dvd_diff
convert h1 using 1
ring
-- d divides (m + n) - (m - n) = 2*n
have h_d_dvd_2n : (d : ℤ) ∣ 2 * n := by
have h1 : (d : ℤ) ∣ (m + n) - (m - n) := dvd_sub hd_dvd_sum hd_dvd_diff
convert h1 using 1
ring
have h_gcd_2m_2n : Int.gcd (2 * m) (2 * n) = 2 * Int.gcd m n := by
rw [Int.gcd_mul_left]
simp
have h_d_dvd_gcd : (d : ℤ) ∣ ((2 * m).gcd (2 * n) : ℤ) :=
Int.dvd_coe_gcd h_d_dvd_2m h_d_dvd_2n
rw [h_gcd_2m_2n, h_coprime] at h_d_dvd_gcd
simp only [Nat.cast_ofNat, mul_one] at h_d_dvd_gcd
have h_d_dvd_gcd_cast : (d : ℕ) ∣ 2 :=
Int.natCast_dvd_natCast.mp h_d_dvd_gcd
have h_factors_of_two: (d : ℕ) ∣ 2 ↔ d = 1 ∨ d = 2 :=
Nat.dvd_prime Nat.prime_two
have h_one_or_two : d = 1 ∨ d = 2 := h_factors_of_two.mp h_d_dvd_gcd_cast
have h_one : d = 1 := by
cases h_one_or_two with
| inl h => exact h
| inr h =>
rw [h] at hd_dvd_sum
have h_2_dvd : (2 : ℤ) ∣ (m + n) := hd_dvd_sum
have : (m + n) % 2 = 0 := Int.emod_eq_zero_of_dvd h_2_dvd
rw [this] at h_sum_odd
norm_num at h_sum_odd
exact h_one