Metamath Proof Explorer


Theorem pythagtriplem16

Description: Lemma for pythagtrip . Show the relationship between M , N , and B . (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Hypotheses pythagtriplem15.1 ⊢ M = C + B + C − B 2
pythagtriplem15.2 ⊢ N = C + B − C − B 2
Assertion pythagtriplem16 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B = 2 ⁢ M ⋅ N

Proof

Step Hyp Ref Expression
1 pythagtriplem15.1 ⊢ M = C + B + C − B 2
2 pythagtriplem15.2 ⊢ N = C + B − C − B 2
3 1 2 oveq12i ⊢ M ⋅ N = C + B + C − B 2 ⁢ C + B − C − B 2
4 nncn ⊢ C ∈ ℕ → C ∈ ℂ
5 nncn ⊢ B ∈ ℕ → B ∈ ℂ
6 addcl ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B ∈ ℂ
7 4 5 6 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B ∈ ℂ
8 7 sqrtcld ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B ∈ ℂ
9 subcl ⊢ C ∈ ℂ ∧ B ∈ ℂ → C − B ∈ ℂ
10 4 5 9 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℂ
11 10 sqrtcld ⊢ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℂ
12 addcl ⊢ C + B ∈ ℂ ∧ C − B ∈ ℂ → C + B + C − B ∈ ℂ
13 8 11 12 syl2anc ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B + C − B ∈ ℂ
14 13 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B + C − B ∈ ℂ
15 14 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B ∈ ℂ
16 subcl ⊢ C + B ∈ ℂ ∧ C − B ∈ ℂ → C + B − C − B ∈ ℂ
17 8 11 16 syl2anc ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B − C − B ∈ ℂ
18 17 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B − C − B ∈ ℂ
19 18 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B − C − B ∈ ℂ
20 2cnne0 ⊢ 2 ∈ ℂ ∧ 2 ≠ 0
21 divmuldiv ⊢ C + B + C − B ∈ ℂ ∧ C + B − C − B ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → C + B + C − B 2 ⁢ C + B − C − B 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
22 20 20 21 mpanr12 ⊢ C + B + C − B ∈ ℂ ∧ C + B − C − B ∈ ℂ → C + B + C − B 2 ⁢ C + B − C − B 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
23 15 19 22 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B 2 ⁢ C + B − C − B 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
24 13 17 mulcld ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B + C − B ⁢ C + B − C − B ∈ ℂ
25 24 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B + C − B ⁢ C + B − C − B ∈ ℂ
26 25 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B ⁢ C + B − C − B ∈ ℂ
27 divdiv1 ⊢ C + B + C − B ⁢ C + B − C − B ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → C + B + C − B ⁢ C + B − C − B 2 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
28 20 20 27 mp3an23 ⊢ C + B + C − B ⁢ C + B − C − B ∈ ℂ → C + B + C − B ⁢ C + B − C − B 2 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
29 26 28 syl ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B ⁢ C + B − C − B 2 2 = C + B + C − B ⁢ C + B − C − B 2 ⋅ 2
30 23 29 eqtr4d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B 2 ⁢ C + B − C − B 2 = C + B + C − B ⁢ C + B − C − B 2 2
31 nnre ⊢ C ∈ ℕ → C ∈ ℝ
32 nnre ⊢ B ∈ ℕ → B ∈ ℝ
33 readdcl ⊢ C ∈ ℝ ∧ B ∈ ℝ → C + B ∈ ℝ
34 31 32 33 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B ∈ ℝ
35 34 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B ∈ ℝ
36 35 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B ∈ ℝ
37 31 adantl ⊢ B ∈ ℕ ∧ C ∈ ℕ → C ∈ ℝ
38 32 adantr ⊢ B ∈ ℕ ∧ C ∈ ℕ → B ∈ ℝ
39 nngt0 ⊢ C ∈ ℕ → 0 < C
40 39 adantl ⊢ B ∈ ℕ ∧ C ∈ ℕ → 0 < C
41 nngt0 ⊢ B ∈ ℕ → 0 < B
42 41 adantr ⊢ B ∈ ℕ ∧ C ∈ ℕ → 0 < B
43 37 38 40 42 addgt0d ⊢ B ∈ ℕ ∧ C ∈ ℕ → 0 < C + B
44 0re ⊢ 0 ∈ ℝ
45 ltle ⊢ 0 ∈ ℝ ∧ C + B ∈ ℝ → 0 < C + B → 0 ≤ C + B
46 44 45 mpan ⊢ C + B ∈ ℝ → 0 < C + B → 0 ≤ C + B
47 34 43 46 sylc ⊢ B ∈ ℕ ∧ C ∈ ℕ → 0 ≤ C + B
48 47 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → 0 ≤ C + B
49 48 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 0 ≤ C + B
50 resqrtth ⊢ C + B ∈ ℝ ∧ 0 ≤ C + B → C + B 2 = C + B
51 36 49 50 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B 2 = C + B
52 resubcl ⊢ C ∈ ℝ ∧ B ∈ ℝ → C − B ∈ ℝ
53 31 32 52 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℝ
54 53 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C − B ∈ ℝ
55 54 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B ∈ ℝ
56 pythagtriplem10 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 → 0 < C − B
57 56 3adant3 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 0 < C − B
58 ltle ⊢ 0 ∈ ℝ ∧ C − B ∈ ℝ → 0 < C − B → 0 ≤ C − B
59 44 58 mpan ⊢ C − B ∈ ℝ → 0 < C − B → 0 ≤ C − B
60 55 57 59 sylc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 0 ≤ C − B
61 resqrtth ⊢ C − B ∈ ℝ ∧ 0 ≤ C − B → C − B 2 = C − B
62 55 60 61 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B 2 = C − B
63 51 62 oveq12d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B 2 − C − B 2 = C + B - C − B
64 63 oveq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B 2 − C − B 2 2 = C + B - C − B 2
65 simp12 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B ∈ ℕ
66 simp13 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C ∈ ℕ
67 65 66 8 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B ∈ ℂ
68 65 66 11 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C − B ∈ ℂ
69 subsq ⊢ C + B ∈ ℂ ∧ C − B ∈ ℂ → C + B 2 − C − B 2 = C + B + C − B ⁢ C + B − C − B
70 67 68 69 syl2anc ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B 2 − C − B 2 = C + B + C − B ⁢ C + B − C − B
71 70 oveq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B 2 − C − B 2 2 = C + B + C − B ⁢ C + B − C − B 2
72 pnncan ⊢ C ∈ ℂ ∧ B ∈ ℂ ∧ B ∈ ℂ → C + B - C − B = B + B
73 72 3anidm23 ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B - C − B = B + B
74 2times ⊢ B ∈ ℂ → 2 ⁢ B = B + B
75 74 adantl ⊢ C ∈ ℂ ∧ B ∈ ℂ → 2 ⁢ B = B + B
76 73 75 eqtr4d ⊢ C ∈ ℂ ∧ B ∈ ℂ → C + B - C − B = 2 ⁢ B
77 4 5 76 syl2anr ⊢ B ∈ ℕ ∧ C ∈ ℕ → C + B - C − B = 2 ⁢ B
78 77 3adant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → C + B - C − B = 2 ⁢ B
79 78 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B - C − B = 2 ⁢ B
80 79 oveq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B - C − B 2 = 2 ⁢ B 2
81 2cn ⊢ 2 ∈ ℂ
82 2ne0 ⊢ 2 ≠ 0
83 divcan3 ⊢ B ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 2 ⁢ B 2 = B
84 81 82 83 mp3an23 ⊢ B ∈ ℂ → 2 ⁢ B 2 = B
85 65 5 84 3syl ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B 2 = B
86 80 85 eqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B - C − B 2 = B
87 64 71 86 3eqtr3d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B ⁢ C + B − C − B 2 = B
88 87 oveq1d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B ⁢ C + B − C − B 2 2 = B 2
89 30 88 eqtrd ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → C + B + C − B 2 ⁢ C + B − C − B 2 = B 2
90 3 89 eqtrid ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → M ⋅ N = B 2
91 90 oveq2d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ M ⋅ N = 2 ⁢ B 2
92 divcan2 ⊢ B ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 2 ⁢ B 2 = B
93 81 82 92 mp3an23 ⊢ B ∈ ℂ → 2 ⁢ B 2 = B
94 5 93 syl ⊢ B ∈ ℕ → 2 ⁢ B 2 = B
95 94 3ad2ant2 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ → 2 ⁢ B 2 = B
96 95 3ad2ant1 ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → 2 ⁢ B 2 = B
97 91 96 eqtr2d ⊢ A ∈ ℕ ∧ B ∈ ℕ ∧ C ∈ ℕ ∧ A 2 + B 2 = C 2 ∧ A gcd B = 1 ∧ ¬ 2 ∥ A → B = 2 ⁢ M ⋅ N