Metamath Proof Explorer


Theorem gausslemma2d

Description: Gauss' Lemma (see also theorem 9.6 in ApostolNT p. 182) for integer 2 : Let p be an odd prime. Let S = {2, 4, 6, ..., p - 1}. Let n denote the number of elements of S whose least positive residue modulo p is greater than p/2. Then ( 2 | p ) = (-1)^n. (Contributed by AV, 14-Jul-2021)

Ref Expression
Hypotheses gausslemma2d.p ⊢ φ → P ∈ ℙ ∖ 2
gausslemma2d.h ⊢ H = P − 1 2
gausslemma2d.r ⊢ R = x ∈ 1 … H ⟼ if x ⋅ 2 < P 2 x ⋅ 2 P − x ⋅ 2
gausslemma2d.m ⊢ M = P 4
gausslemma2d.n ⊢ N = H − M
Assertion gausslemma2d ⊢ φ → 2 / L P = − 1 N

Proof

Step Hyp Ref Expression
1 gausslemma2d.p ⊢ φ → P ∈ ℙ ∖ 2
2 gausslemma2d.h ⊢ H = P − 1 2
3 gausslemma2d.r ⊢ R = x ∈ 1 … H ⟼ if x ⋅ 2 < P 2 x ⋅ 2 P − x ⋅ 2
4 gausslemma2d.m ⊢ M = P 4
5 gausslemma2d.n ⊢ N = H − M
6 1 2 3 4 5 gausslemma2dlem7 ⊢ φ → − 1 N ⁢ 2 H mod P = 1
7 eldifi ⊢ P ∈ ℙ ∖ 2 → P ∈ ℙ
8 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
9 8 nnred ⊢ P ∈ ℙ → P ∈ ℝ
10 prmgt1 ⊢ P ∈ ℙ → 1 < P
11 9 10 jca ⊢ P ∈ ℙ → P ∈ ℝ ∧ 1 < P
12 1mod ⊢ P ∈ ℝ ∧ 1 < P → 1 mod P = 1
13 1 7 11 12 4syl ⊢ φ → 1 mod P = 1
14 13 eqcomd ⊢ φ → 1 = 1 mod P
15 14 eqeq2d ⊢ φ → − 1 N ⁢ 2 H mod P = 1 ↔ − 1 N ⁢ 2 H mod P = 1 mod P
16 neg1z ⊢ − 1 ∈ ℤ
17 1 4 2 5 gausslemma2dlem0h ⊢ φ → N ∈ ℕ 0
18 zexpcl ⊢ − 1 ∈ ℤ ∧ N ∈ ℕ 0 → − 1 N ∈ ℤ
19 16 17 18 sylancr ⊢ φ → − 1 N ∈ ℤ
20 2nn ⊢ 2 ∈ ℕ
21 20 a1i ⊢ φ → 2 ∈ ℕ
22 1 2 gausslemma2dlem0b ⊢ φ → H ∈ ℕ
23 22 nnnn0d ⊢ φ → H ∈ ℕ 0
24 21 23 nnexpcld ⊢ φ → 2 H ∈ ℕ
25 24 nnzd ⊢ φ → 2 H ∈ ℤ
26 19 25 zmulcld ⊢ φ → − 1 N ⁢ 2 H ∈ ℤ
27 26 zred ⊢ φ → − 1 N ⁢ 2 H ∈ ℝ
28 1red ⊢ φ → 1 ∈ ℝ
29 27 28 jca ⊢ φ → − 1 N ⁢ 2 H ∈ ℝ ∧ 1 ∈ ℝ
30 29 adantr ⊢ φ ∧ − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ⁢ 2 H ∈ ℝ ∧ 1 ∈ ℝ
31 1 gausslemma2dlem0a ⊢ φ → P ∈ ℕ
32 31 nnrpd ⊢ φ → P ∈ ℝ +
33 19 32 jca ⊢ φ → − 1 N ∈ ℤ ∧ P ∈ ℝ +
34 33 adantr ⊢ φ ∧ − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ∈ ℤ ∧ P ∈ ℝ +
35 simpr ⊢ φ ∧ − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ⁢ 2 H mod P = 1 mod P
36 modmul1 ⊢ − 1 N ⁢ 2 H ∈ ℝ ∧ 1 ∈ ℝ ∧ − 1 N ∈ ℤ ∧ P ∈ ℝ + ∧ − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 1 ⁢ − 1 N mod P
37 30 34 35 36 syl3anc ⊢ φ ∧ − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 1 ⁢ − 1 N mod P
38 37 ex ⊢ φ → − 1 N ⁢ 2 H mod P = 1 mod P → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 1 ⁢ − 1 N mod P
39 19 zcnd ⊢ φ → − 1 N ∈ ℂ
40 24 nncnd ⊢ φ → 2 H ∈ ℂ
41 39 40 39 mul32d ⊢ φ → − 1 N ⁢ 2 H ⁢ − 1 N = − 1 N ⁢ − 1 N ⁢ 2 H
42 17 nn0cnd ⊢ φ → N ∈ ℂ
43 42 2timesd ⊢ φ → 2 ⋅ N = N + N
44 43 eqcomd ⊢ φ → N + N = 2 ⋅ N
45 44 oveq2d ⊢ φ → − 1 N + N = − 1 2 ⋅ N
46 neg1cn ⊢ − 1 ∈ ℂ
47 46 a1i ⊢ φ → − 1 ∈ ℂ
48 47 17 17 expaddd ⊢ φ → − 1 N + N = − 1 N ⁢ − 1 N
49 17 nn0zd ⊢ φ → N ∈ ℤ
50 m1expeven ⊢ N ∈ ℤ → − 1 2 ⋅ N = 1
51 49 50 syl ⊢ φ → − 1 2 ⋅ N = 1
52 45 48 51 3eqtr3d ⊢ φ → − 1 N ⁢ − 1 N = 1
53 52 oveq1d ⊢ φ → − 1 N ⁢ − 1 N ⁢ 2 H = 1 ⁢ 2 H
54 40 mullidd ⊢ φ → 1 ⁢ 2 H = 2 H
55 41 53 54 3eqtrd ⊢ φ → − 1 N ⁢ 2 H ⁢ − 1 N = 2 H
56 55 oveq1d ⊢ φ → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 2 H mod P
57 39 mullidd ⊢ φ → 1 ⁢ − 1 N = − 1 N
58 57 oveq1d ⊢ φ → 1 ⁢ − 1 N mod P = − 1 N mod P
59 56 58 eqeq12d ⊢ φ → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 1 ⁢ − 1 N mod P ↔ 2 H mod P = − 1 N mod P
60 2 oveq2i ⊢ 2 H = 2 P − 1 2
61 60 oveq1i ⊢ 2 H mod P = 2 P − 1 2 mod P
62 61 eqeq1i ⊢ 2 H mod P = − 1 N mod P ↔ 2 P − 1 2 mod P = − 1 N mod P
63 2z ⊢ 2 ∈ ℤ
64 lgsvalmod ⊢ 2 ∈ ℤ ∧ P ∈ ℙ ∖ 2 → 2 / L P mod P = 2 P − 1 2 mod P
65 63 1 64 sylancr ⊢ φ → 2 / L P mod P = 2 P − 1 2 mod P
66 65 eqcomd ⊢ φ → 2 P − 1 2 mod P = 2 / L P mod P
67 66 eqeq1d ⊢ φ → 2 P − 1 2 mod P = − 1 N mod P ↔ 2 / L P mod P = − 1 N mod P
68 1 4 2 5 gausslemma2dlem0i ⊢ φ → 2 / L P mod P = − 1 N mod P → 2 / L P = − 1 N
69 67 68 sylbid ⊢ φ → 2 P − 1 2 mod P = − 1 N mod P → 2 / L P = − 1 N
70 62 69 biimtrid ⊢ φ → 2 H mod P = − 1 N mod P → 2 / L P = − 1 N
71 59 70 sylbid ⊢ φ → − 1 N ⁢ 2 H ⁢ − 1 N mod P = 1 ⁢ − 1 N mod P → 2 / L P = − 1 N
72 38 71 syld ⊢ φ → − 1 N ⁢ 2 H mod P = 1 mod P → 2 / L P = − 1 N
73 15 72 sylbid ⊢ φ → − 1 N ⁢ 2 H mod P = 1 → 2 / L P = − 1 N
74 6 73 mpd ⊢ φ → 2 / L P = − 1 N