Metamath Proof Explorer


Theorem flt4lem5e

Description: Satisfy the hypotheses of flt4lem4 . (Contributed by SN, 23-Aug-2024)

Ref Expression
Hypotheses flt4lem5a.m ⊢ M = C + B 2 + C − B 2 2
flt4lem5a.n ⊢ N = C + B 2 − C − B 2 2
flt4lem5a.r ⊢ R = M + N + M − N 2
flt4lem5a.s ⊢ S = M + N − M − N 2
flt4lem5a.a ⊢ φ → A ∈ ℕ
flt4lem5a.b ⊢ φ → B ∈ ℕ
flt4lem5a.c ⊢ φ → C ∈ ℕ
flt4lem5a.1 ⊢ φ → ¬ 2 ∥ A
flt4lem5a.2 ⊢ φ → A gcd C = 1
flt4lem5a.3 ⊢ φ → A 4 + B 4 = C 2
Assertion flt4lem5e ⊢ φ → R gcd S = 1 ∧ R gcd M = 1 ∧ S gcd M = 1 ∧ R ∈ ℕ ∧ S ∈ ℕ ∧ M ∈ ℕ ∧ M ⁢ R ⁢ S = B 2 2 ∧ B 2 ∈ ℕ

Proof

Step Hyp Ref Expression
1 flt4lem5a.m ⊢ M = C + B 2 + C − B 2 2
2 flt4lem5a.n ⊢ N = C + B 2 − C − B 2 2
3 flt4lem5a.r ⊢ R = M + N + M − N 2
4 flt4lem5a.s ⊢ S = M + N − M − N 2
5 flt4lem5a.a ⊢ φ → A ∈ ℕ
6 flt4lem5a.b ⊢ φ → B ∈ ℕ
7 flt4lem5a.c ⊢ φ → C ∈ ℕ
8 flt4lem5a.1 ⊢ φ → ¬ 2 ∥ A
9 flt4lem5a.2 ⊢ φ → A gcd C = 1
10 flt4lem5a.3 ⊢ φ → A 4 + B 4 = C 2
11 5 nnsqcld ⊢ φ → A 2 ∈ ℕ
12 6 nnsqcld ⊢ φ → B 2 ∈ ℕ
13 2prm ⊢ 2 ∈ ℙ
14 5 nnzd ⊢ φ → A ∈ ℤ
15 prmdvdssq ⊢ 2 ∈ ℙ ∧ A ∈ ℤ → 2 ∥ A ↔ 2 ∥ A 2
16 13 14 15 sylancr ⊢ φ → 2 ∥ A ↔ 2 ∥ A 2
17 8 16 mtbid ⊢ φ → ¬ 2 ∥ A 2
18 2nn ⊢ 2 ∈ ℕ
19 18 a1i ⊢ φ → 2 ∈ ℕ
20 rplpwr ⊢ A ∈ ℕ ∧ C ∈ ℕ ∧ 2 ∈ ℕ → A gcd C = 1 → A 2 gcd C = 1
21 5 7 19 20 syl3anc ⊢ φ → A gcd C = 1 → A 2 gcd C = 1
22 9 21 mpd ⊢ φ → A 2 gcd C = 1
23 5 nncnd ⊢ φ → A ∈ ℂ
24 23 exp4sqsq ⊢ φ → A 4 = A 2 2
25 6 nncnd ⊢ φ → B ∈ ℂ
26 25 exp4sqsq ⊢ φ → B 4 = B 2 2
27 24 26 oveq12d ⊢ φ → A 4 + B 4 = A 2 2 + B 2 2
28 27 10 eqtr3d ⊢ φ → A 2 2 + B 2 2 = C 2
29 11 12 7 17 22 28 flt4lem1 ⊢ φ → A 2 ∈ ℕ ∧ B 2 ∈ ℕ ∧ C ∈ ℕ ∧ A 2 2 + B 2 2 = C 2 ∧ A 2 gcd B 2 = 1 ∧ ¬ 2 ∥ A 2
30 2 pythagtriplem13 ⊢ A 2 ∈ ℕ ∧ B 2 ∈ ℕ ∧ C ∈ ℕ ∧ A 2 2 + B 2 2 = C 2 ∧ A 2 gcd B 2 = 1 ∧ ¬ 2 ∥ A 2 → N ∈ ℕ
31 29 30 syl ⊢ φ → N ∈ ℕ
32 1 pythagtriplem11 ⊢ A 2 ∈ ℕ ∧ B 2 ∈ ℕ ∧ C ∈ ℕ ∧ A 2 2 + B 2 2 = C 2 ∧ A 2 gcd B 2 = 1 ∧ ¬ 2 ∥ A 2 → M ∈ ℕ
33 29 32 syl ⊢ φ → M ∈ ℕ
34 1 2 3 4 5 6 7 8 9 10 flt4lem5a ⊢ φ → A 2 + N 2 = M 2
35 31 nnzd ⊢ φ → N ∈ ℤ
36 14 35 gcdcomd ⊢ φ → A gcd N = N gcd A
37 33 nnzd ⊢ φ → M ∈ ℤ
38 35 37 gcdcomd ⊢ φ → N gcd M = M gcd N
39 1 2 flt4lem5 ⊢ A 2 ∈ ℕ ∧ B 2 ∈ ℕ ∧ C ∈ ℕ ∧ A 2 2 + B 2 2 = C 2 ∧ A 2 gcd B 2 = 1 ∧ ¬ 2 ∥ A 2 → M gcd N = 1
40 29 39 syl ⊢ φ → M gcd N = 1
41 38 40 eqtrd ⊢ φ → N gcd M = 1
42 31 nnsqcld ⊢ φ → N 2 ∈ ℕ
43 42 nncnd ⊢ φ → N 2 ∈ ℂ
44 11 nncnd ⊢ φ → A 2 ∈ ℂ
45 43 44 addcomd ⊢ φ → N 2 + A 2 = A 2 + N 2
46 45 34 eqtrd ⊢ φ → N 2 + A 2 = M 2
47 31 5 33 41 46 fltabcoprm ⊢ φ → N gcd A = 1
48 36 47 eqtrd ⊢ φ → A gcd N = 1
49 3 4 flt4lem5 ⊢ A ∈ ℕ ∧ N ∈ ℕ ∧ M ∈ ℕ ∧ A 2 + N 2 = M 2 ∧ A gcd N = 1 ∧ ¬ 2 ∥ A → R gcd S = 1
50 5 31 33 34 48 8 49 syl312anc ⊢ φ → R gcd S = 1
51 3 pythagtriplem11 ⊢ A ∈ ℕ ∧ N ∈ ℕ ∧ M ∈ ℕ ∧ A 2 + N 2 = M 2 ∧ A gcd N = 1 ∧ ¬ 2 ∥ A → R ∈ ℕ
52 5 31 33 34 48 8 51 syl312anc ⊢ φ → R ∈ ℕ
53 4 pythagtriplem13 ⊢ A ∈ ℕ ∧ N ∈ ℕ ∧ M ∈ ℕ ∧ A 2 + N 2 = M 2 ∧ A gcd N = 1 ∧ ¬ 2 ∥ A → S ∈ ℕ
54 5 31 33 34 48 8 53 syl312anc ⊢ φ → S ∈ ℕ
55 1 2 3 4 5 6 7 8 9 10 flt4lem5d ⊢ φ → M = R 2 + S 2
56 33 52 54 55 50 flt4lem5elem ⊢ φ → R gcd M = 1 ∧ S gcd M = 1
57 3anass ⊢ R gcd S = 1 ∧ R gcd M = 1 ∧ S gcd M = 1 ↔ R gcd S = 1 ∧ R gcd M = 1 ∧ S gcd M = 1
58 50 56 57 sylanbrc ⊢ φ → R gcd S = 1 ∧ R gcd M = 1 ∧ S gcd M = 1
59 52 54 33 3jca ⊢ φ → R ∈ ℕ ∧ S ∈ ℕ ∧ M ∈ ℕ
60 sq2 ⊢ 2 2 = 4
61 4cn ⊢ 4 ∈ ℂ
62 60 61 eqeltri ⊢ 2 2 ∈ ℂ
63 62 a1i ⊢ φ → 2 2 ∈ ℂ
64 52 54 nnmulcld ⊢ φ → R ⁢ S ∈ ℕ
65 33 64 nnmulcld ⊢ φ → M ⁢ R ⁢ S ∈ ℕ
66 65 nncnd ⊢ φ → M ⁢ R ⁢ S ∈ ℂ
67 4ne0 ⊢ 4 ≠ 0
68 60 67 eqnetri ⊢ 2 2 ≠ 0
69 68 a1i ⊢ φ → 2 2 ≠ 0
70 2cn ⊢ 2 ∈ ℂ
71 70 sqvali ⊢ 2 2 = 2 ⋅ 2
72 71 oveq1i ⊢ 2 2 ⁢ M ⁢ R ⁢ S = 2 ⋅ 2 ⁢ M ⁢ R ⁢ S
73 2cnd ⊢ φ → 2 ∈ ℂ
74 33 nncnd ⊢ φ → M ∈ ℂ
75 64 nncnd ⊢ φ → R ⁢ S ∈ ℂ
76 73 73 74 75 mul4d ⊢ φ → 2 ⋅ 2 ⁢ M ⁢ R ⁢ S = 2 ⋅ M ⁢ 2 ⁢ R ⁢ S
77 1 2 3 4 5 6 7 8 9 10 flt4lem5c ⊢ φ → N = 2 ⁢ R ⁢ S
78 77 31 eqeltrrd ⊢ φ → 2 ⁢ R ⁢ S ∈ ℕ
79 78 nncnd ⊢ φ → 2 ⁢ R ⁢ S ∈ ℂ
80 73 74 79 mulassd ⊢ φ → 2 ⋅ M ⁢ 2 ⁢ R ⁢ S = 2 ⁢ M ⁢ 2 ⁢ R ⁢ S
81 77 eqcomd ⊢ φ → 2 ⁢ R ⁢ S = N
82 81 oveq2d ⊢ φ → M ⁢ 2 ⁢ R ⁢ S = M ⋅ N
83 82 oveq2d ⊢ φ → 2 ⁢ M ⁢ 2 ⁢ R ⁢ S = 2 ⁢ M ⋅ N
84 80 83 eqtrd ⊢ φ → 2 ⋅ M ⁢ 2 ⁢ R ⁢ S = 2 ⁢ M ⋅ N
85 1 2 3 4 5 6 7 8 9 10 flt4lem5b ⊢ φ → 2 ⁢ M ⋅ N = B 2
86 76 84 85 3eqtrd ⊢ φ → 2 ⋅ 2 ⁢ M ⁢ R ⁢ S = B 2
87 72 86 eqtrid ⊢ φ → 2 2 ⁢ M ⁢ R ⁢ S = B 2
88 63 66 69 87 mvllmuld ⊢ φ → M ⁢ R ⁢ S = B 2 2 2
89 2ne0 ⊢ 2 ≠ 0
90 89 a1i ⊢ φ → 2 ≠ 0
91 25 73 90 sqdivd ⊢ φ → B 2 2 = B 2 2 2
92 88 91 eqtr4d ⊢ φ → M ⁢ R ⁢ S = B 2 2
93 65 nnzd ⊢ φ → M ⁢ R ⁢ S ∈ ℤ
94 92 93 eqeltrrd ⊢ φ → B 2 2 ∈ ℤ
95 6 nnzd ⊢ φ → B ∈ ℤ
96 znq ⊢ B ∈ ℤ ∧ 2 ∈ ℕ → B 2 ∈ ℚ
97 95 18 96 sylancl ⊢ φ → B 2 ∈ ℚ
98 6 nngt0d ⊢ φ → 0 < B
99 6 nnred ⊢ φ → B ∈ ℝ
100 halfpos2 ⊢ B ∈ ℝ → 0 < B ↔ 0 < B 2
101 99 100 syl ⊢ φ → 0 < B ↔ 0 < B 2
102 98 101 mpbid ⊢ φ → 0 < B 2
103 94 97 102 posqsqznn ⊢ φ → B 2 ∈ ℕ
104 92 103 jca ⊢ φ → M ⁢ R ⁢ S = B 2 2 ∧ B 2 ∈ ℕ
105 58 59 104 3jca ⊢ φ → R gcd S = 1 ∧ R gcd M = 1 ∧ S gcd M = 1 ∧ R ∈ ℕ ∧ S ∈ ℕ ∧ M ∈ ℕ ∧ M ⁢ R ⁢ S = B 2 2 ∧ B 2 ∈ ℕ