Metamath Proof Explorer


Theorem 2sqlem4

Description: Lemma for 2sqlem5 . (Contributed by Mario Carneiro, 20-Jun-2015)

Ref Expression
Hypotheses 2sq.1 ⊢ S = ran ⁡ w ∈ ℤ i ⟼ w 2
2sqlem5.1 ⊢ φ → N ∈ ℕ
2sqlem5.2 ⊢ φ → P ∈ ℙ
2sqlem4.3 ⊢ φ → A ∈ ℤ
2sqlem4.4 ⊢ φ → B ∈ ℤ
2sqlem4.5 ⊢ φ → C ∈ ℤ
2sqlem4.6 ⊢ φ → D ∈ ℤ
2sqlem4.7 ⊢ φ → N ⁢ P = A 2 + B 2
2sqlem4.8 ⊢ φ → P = C 2 + D 2
Assertion 2sqlem4 ⊢ φ → N ∈ S

Proof

Step Hyp Ref Expression
1 2sq.1 ⊢ S = ran ⁡ w ∈ ℤ i ⟼ w 2
2 2sqlem5.1 ⊢ φ → N ∈ ℕ
3 2sqlem5.2 ⊢ φ → P ∈ ℙ
4 2sqlem4.3 ⊢ φ → A ∈ ℤ
5 2sqlem4.4 ⊢ φ → B ∈ ℤ
6 2sqlem4.5 ⊢ φ → C ∈ ℤ
7 2sqlem4.6 ⊢ φ → D ∈ ℤ
8 2sqlem4.7 ⊢ φ → N ⁢ P = A 2 + B 2
9 2sqlem4.8 ⊢ φ → P = C 2 + D 2
10 2 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → N ∈ ℕ
11 3 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → P ∈ ℙ
12 4 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → A ∈ ℤ
13 5 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → B ∈ ℤ
14 6 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → C ∈ ℤ
15 7 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → D ∈ ℤ
16 8 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → N ⁢ P = A 2 + B 2
17 9 adantr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → P = C 2 + D 2
18 simpr ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → P ∥ C ⁢ B + A ⁢ D
19 1 10 11 12 13 14 15 16 17 18 2sqlem3 ⊢ φ ∧ P ∥ C ⁢ B + A ⁢ D → N ∈ S
20 2 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → N ∈ ℕ
21 3 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → P ∈ ℙ
22 4 znegcld ⊢ φ → − A ∈ ℤ
23 22 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → − A ∈ ℤ
24 5 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → B ∈ ℤ
25 6 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → C ∈ ℤ
26 7 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → D ∈ ℤ
27 4 zcnd ⊢ φ → A ∈ ℂ
28 sqneg ⊢ A ∈ ℂ → − A 2 = A 2
29 27 28 syl ⊢ φ → − A 2 = A 2
30 29 oveq1d ⊢ φ → − A 2 + B 2 = A 2 + B 2
31 8 30 eqtr4d ⊢ φ → N ⁢ P = − A 2 + B 2
32 31 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → N ⁢ P = − A 2 + B 2
33 9 adantr ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → P = C 2 + D 2
34 7 zcnd ⊢ φ → D ∈ ℂ
35 27 34 mulneg1d ⊢ φ → − A ⁢ D = − A ⁢ D
36 35 oveq2d ⊢ φ → C ⁢ B + − A ⁢ D = C ⁢ B + − A ⁢ D
37 6 5 zmulcld ⊢ φ → C ⁢ B ∈ ℤ
38 37 zcnd ⊢ φ → C ⁢ B ∈ ℂ
39 4 7 zmulcld ⊢ φ → A ⁢ D ∈ ℤ
40 39 zcnd ⊢ φ → A ⁢ D ∈ ℂ
41 38 40 negsubd ⊢ φ → C ⁢ B + − A ⁢ D = C ⁢ B − A ⁢ D
42 36 41 eqtrd ⊢ φ → C ⁢ B + − A ⁢ D = C ⁢ B − A ⁢ D
43 42 breq2d ⊢ φ → P ∥ C ⁢ B + − A ⁢ D ↔ P ∥ C ⁢ B − A ⁢ D
44 43 biimpar ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → P ∥ C ⁢ B + − A ⁢ D
45 1 20 21 23 24 25 26 32 33 44 2sqlem3 ⊢ φ ∧ P ∥ C ⁢ B − A ⁢ D → N ∈ S
46 prmz ⊢ P ∈ ℙ → P ∈ ℤ
47 3 46 syl ⊢ φ → P ∈ ℤ
48 zsqcl ⊢ C ∈ ℤ → C 2 ∈ ℤ
49 6 48 syl ⊢ φ → C 2 ∈ ℤ
50 2 nnzd ⊢ φ → N ∈ ℤ
51 49 50 zmulcld ⊢ φ → C 2 ⋅ N ∈ ℤ
52 zsqcl ⊢ A ∈ ℤ → A 2 ∈ ℤ
53 4 52 syl ⊢ φ → A 2 ∈ ℤ
54 51 53 zsubcld ⊢ φ → C 2 ⋅ N − A 2 ∈ ℤ
55 dvdsmul1 ⊢ P ∈ ℤ ∧ C 2 ⋅ N − A 2 ∈ ℤ → P ∥ P ⁢ C 2 ⋅ N − A 2
56 47 54 55 syl2anc ⊢ φ → P ∥ P ⁢ C 2 ⋅ N − A 2
57 6 4 zmulcld ⊢ φ → C ⁢ A ∈ ℤ
58 57 zcnd ⊢ φ → C ⁢ A ∈ ℂ
59 58 sqcld ⊢ φ → C ⁢ A 2 ∈ ℂ
60 38 sqcld ⊢ φ → C ⁢ B 2 ∈ ℂ
61 40 sqcld ⊢ φ → A ⁢ D 2 ∈ ℂ
62 59 60 61 pnpcand ⊢ φ → C ⁢ A 2 + C ⁢ B 2 - C ⁢ A 2 + A ⁢ D 2 = C ⁢ B 2 − A ⁢ D 2
63 6 zcnd ⊢ φ → C ∈ ℂ
64 63 27 sqmuld ⊢ φ → C ⁢ A 2 = C 2 ⁢ A 2
65 5 zcnd ⊢ φ → B ∈ ℂ
66 63 65 sqmuld ⊢ φ → C ⁢ B 2 = C 2 ⁢ B 2
67 64 66 oveq12d ⊢ φ → C ⁢ A 2 + C ⁢ B 2 = C 2 ⁢ A 2 + C 2 ⁢ B 2
68 63 sqcld ⊢ φ → C 2 ∈ ℂ
69 53 zcnd ⊢ φ → A 2 ∈ ℂ
70 65 sqcld ⊢ φ → B 2 ∈ ℂ
71 68 69 70 adddid ⊢ φ → C 2 ⁢ A 2 + B 2 = C 2 ⁢ A 2 + C 2 ⁢ B 2
72 67 71 eqtr4d ⊢ φ → C ⁢ A 2 + C ⁢ B 2 = C 2 ⁢ A 2 + B 2
73 2 nncnd ⊢ φ → N ∈ ℂ
74 47 zcnd ⊢ φ → P ∈ ℂ
75 73 74 mulcomd ⊢ φ → N ⁢ P = P ⋅ N
76 8 75 eqtr3d ⊢ φ → A 2 + B 2 = P ⋅ N
77 76 oveq2d ⊢ φ → C 2 ⁢ A 2 + B 2 = C 2 ⁢ P ⋅ N
78 68 74 73 mul12d ⊢ φ → C 2 ⁢ P ⋅ N = P ⁢ C 2 ⋅ N
79 77 78 eqtrd ⊢ φ → C 2 ⁢ A 2 + B 2 = P ⁢ C 2 ⋅ N
80 72 79 eqtrd ⊢ φ → C ⁢ A 2 + C ⁢ B 2 = P ⁢ C 2 ⋅ N
81 27 34 sqmuld ⊢ φ → A ⁢ D 2 = A 2 ⁢ D 2
82 34 sqcld ⊢ φ → D 2 ∈ ℂ
83 69 82 mulcomd ⊢ φ → A 2 ⁢ D 2 = D 2 ⁢ A 2
84 81 83 eqtrd ⊢ φ → A ⁢ D 2 = D 2 ⁢ A 2
85 64 84 oveq12d ⊢ φ → C ⁢ A 2 + A ⁢ D 2 = C 2 ⁢ A 2 + D 2 ⁢ A 2
86 49 zcnd ⊢ φ → C 2 ∈ ℂ
87 86 82 69 adddird ⊢ φ → C 2 + D 2 ⁢ A 2 = C 2 ⁢ A 2 + D 2 ⁢ A 2
88 85 87 eqtr4d ⊢ φ → C ⁢ A 2 + A ⁢ D 2 = C 2 + D 2 ⁢ A 2
89 9 oveq1d ⊢ φ → P ⁢ A 2 = C 2 + D 2 ⁢ A 2
90 88 89 eqtr4d ⊢ φ → C ⁢ A 2 + A ⁢ D 2 = P ⁢ A 2
91 80 90 oveq12d ⊢ φ → C ⁢ A 2 + C ⁢ B 2 - C ⁢ A 2 + A ⁢ D 2 = P ⁢ C 2 ⋅ N − P ⁢ A 2
92 51 zcnd ⊢ φ → C 2 ⋅ N ∈ ℂ
93 74 92 69 subdid ⊢ φ → P ⁢ C 2 ⋅ N − A 2 = P ⁢ C 2 ⋅ N − P ⁢ A 2
94 91 93 eqtr4d ⊢ φ → C ⁢ A 2 + C ⁢ B 2 - C ⁢ A 2 + A ⁢ D 2 = P ⁢ C 2 ⋅ N − A 2
95 62 94 eqtr3d ⊢ φ → C ⁢ B 2 − A ⁢ D 2 = P ⁢ C 2 ⋅ N − A 2
96 subsq ⊢ C ⁢ B ∈ ℂ ∧ A ⁢ D ∈ ℂ → C ⁢ B 2 − A ⁢ D 2 = C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D
97 38 40 96 syl2anc ⊢ φ → C ⁢ B 2 − A ⁢ D 2 = C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D
98 95 97 eqtr3d ⊢ φ → P ⁢ C 2 ⋅ N − A 2 = C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D
99 56 98 breqtrd ⊢ φ → P ∥ C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D
100 37 39 zaddcld ⊢ φ → C ⁢ B + A ⁢ D ∈ ℤ
101 37 39 zsubcld ⊢ φ → C ⁢ B − A ⁢ D ∈ ℤ
102 euclemma ⊢ P ∈ ℙ ∧ C ⁢ B + A ⁢ D ∈ ℤ ∧ C ⁢ B − A ⁢ D ∈ ℤ → P ∥ C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D ↔ P ∥ C ⁢ B + A ⁢ D ∨ P ∥ C ⁢ B − A ⁢ D
103 3 100 101 102 syl3anc ⊢ φ → P ∥ C ⁢ B + A ⁢ D ⁢ C ⁢ B − A ⁢ D ↔ P ∥ C ⁢ B + A ⁢ D ∨ P ∥ C ⁢ B − A ⁢ D
104 99 103 mpbid ⊢ φ → P ∥ C ⁢ B + A ⁢ D ∨ P ∥ C ⁢ B − A ⁢ D
105 19 45 104 mpjaodan ⊢ φ → N ∈ S