Metamath Proof Explorer


Theorem pythagtriplem19

Description: Lemma for pythagtrip . Introduce k and remove the relative primality requirement. (Contributed by Scott Fenton, 18-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion pythagtriplem19 ABCA2+B2=C2¬2AAgcdBnmkA=km2n2B=k2mnC=km2+n2

Proof

Step Hyp Ref Expression
1 gcdnncl ABAgcdB
2 1 3adant3 ABCAgcdB
3 2 3ad2ant1 ABCA2+B2=C2¬2AAgcdBAgcdB
4 nnz AA
5 nnz BB
6 gcddvds ABAgcdBAAgcdBB
7 4 5 6 syl2an ABAgcdBAAgcdBB
8 7 3adant3 ABCAgcdBAAgcdBB
9 8 simpld ABCAgcdBA
10 2 nnzd ABCAgcdB
11 2 nnne0d ABCAgcdB0
12 4 3ad2ant1 ABCA
13 dvdsval2 AgcdBAgcdB0AAgcdBAAAgcdB
14 10 11 12 13 syl3anc ABCAgcdBAAAgcdB
15 9 14 mpbid ABCAAgcdB
16 nnre AA
17 16 3ad2ant1 ABCA
18 2 nnred ABCAgcdB
19 nngt0 A0<A
20 19 3ad2ant1 ABC0<A
21 2 nngt0d ABC0<AgcdB
22 17 18 20 21 divgt0d ABC0<AAgcdB
23 elnnz AAgcdBAAgcdB0<AAgcdB
24 15 22 23 sylanbrc ABCAAgcdB
25 24 3ad2ant1 ABCA2+B2=C2¬2AAgcdBAAgcdB
26 8 simprd ABCAgcdBB
27 5 3ad2ant2 ABCB
28 dvdsval2 AgcdBAgcdB0BAgcdBBBAgcdB
29 10 11 27 28 syl3anc ABCAgcdBBBAgcdB
30 26 29 mpbid ABCBAgcdB
31 nnre BB
32 31 3ad2ant2 ABCB
33 nngt0 B0<B
34 33 3ad2ant2 ABC0<B
35 32 18 34 21 divgt0d ABC0<BAgcdB
36 elnnz BAgcdBBAgcdB0<BAgcdB
37 30 35 36 sylanbrc ABCBAgcdB
38 37 3ad2ant1 ABCA2+B2=C2¬2AAgcdBBAgcdB
39 dvdssq AgcdBAAgcdBAAgcdB2A2
40 10 12 39 syl2anc ABCAgcdBAAgcdB2A2
41 dvdssq AgcdBBAgcdBBAgcdB2B2
42 10 27 41 syl2anc ABCAgcdBBAgcdB2B2
43 40 42 anbi12d ABCAgcdBAAgcdBBAgcdB2A2AgcdB2B2
44 8 43 mpbid ABCAgcdB2A2AgcdB2B2
45 2 nnsqcld ABCAgcdB2
46 45 nnzd ABCAgcdB2
47 nnsqcl AA2
48 47 3ad2ant1 ABCA2
49 48 nnzd ABCA2
50 nnsqcl BB2
51 50 3ad2ant2 ABCB2
52 51 nnzd ABCB2
53 dvds2add AgcdB2A2B2AgcdB2A2AgcdB2B2AgcdB2A2+B2
54 46 49 52 53 syl3anc ABCAgcdB2A2AgcdB2B2AgcdB2A2+B2
55 44 54 mpd ABCAgcdB2A2+B2
56 55 adantr ABCA2+B2=C2AgcdB2A2+B2
57 simpr ABCA2+B2=C2A2+B2=C2
58 56 57 breqtrd ABCA2+B2=C2AgcdB2C2
59 nnz CC
60 59 3ad2ant3 ABCC
61 dvdssq AgcdBCAgcdBCAgcdB2C2
62 10 60 61 syl2anc ABCAgcdBCAgcdB2C2
63 62 adantr ABCA2+B2=C2AgcdBCAgcdB2C2
64 58 63 mpbird ABCA2+B2=C2AgcdBC
65 dvdsval2 AgcdBAgcdB0CAgcdBCCAgcdB
66 10 11 60 65 syl3anc ABCAgcdBCCAgcdB
67 66 adantr ABCA2+B2=C2AgcdBCCAgcdB
68 64 67 mpbid ABCA2+B2=C2CAgcdB
69 nnre CC
70 69 3ad2ant3 ABCC
71 nngt0 C0<C
72 71 3ad2ant3 ABC0<C
73 70 18 72 21 divgt0d ABC0<CAgcdB
74 73 adantr ABCA2+B2=C20<CAgcdB
75 elnnz CAgcdBCAgcdB0<CAgcdB
76 68 74 75 sylanbrc ABCA2+B2=C2CAgcdB
77 76 3adant3 ABCA2+B2=C2¬2AAgcdBCAgcdB
78 48 nncnd ABCA2
79 51 nncnd ABCB2
80 45 nncnd ABCAgcdB2
81 45 nnne0d ABCAgcdB20
82 78 79 80 81 divdird ABCA2+B2AgcdB2=A2AgcdB2+B2AgcdB2
83 82 3ad2ant1 ABCA2+B2=C2¬2AAgcdBA2+B2AgcdB2=A2AgcdB2+B2AgcdB2
84 nncn CC
85 84 3ad2ant3 ABCC
86 2 nncnd ABCAgcdB
87 85 86 11 sqdivd ABCCAgcdB2=C2AgcdB2
88 87 3ad2ant1 ABCA2+B2=C2¬2AAgcdBCAgcdB2=C2AgcdB2
89 oveq1 A2+B2=C2A2+B2AgcdB2=C2AgcdB2
90 89 3ad2ant2 ABCA2+B2=C2¬2AAgcdBA2+B2AgcdB2=C2AgcdB2
91 88 90 eqtr4d ABCA2+B2=C2¬2AAgcdBCAgcdB2=A2+B2AgcdB2
92 nncn AA
93 92 3ad2ant1 ABCA
94 93 86 11 sqdivd ABCAAgcdB2=A2AgcdB2
95 nncn BB
96 95 3ad2ant2 ABCB
97 96 86 11 sqdivd ABCBAgcdB2=B2AgcdB2
98 94 97 oveq12d ABCAAgcdB2+BAgcdB2=A2AgcdB2+B2AgcdB2
99 98 3ad2ant1 ABCA2+B2=C2¬2AAgcdBAAgcdB2+BAgcdB2=A2AgcdB2+B2AgcdB2
100 83 91 99 3eqtr4rd ABCA2+B2=C2¬2AAgcdBAAgcdB2+BAgcdB2=CAgcdB2
101 gcddiv ABAgcdBAgcdBAAgcdBBAgcdBAgcdB=AAgcdBgcdBAgcdB
102 12 27 2 8 101 syl31anc ABCAgcdBAgcdB=AAgcdBgcdBAgcdB
103 86 11 dividd ABCAgcdBAgcdB=1
104 102 103 eqtr3d ABCAAgcdBgcdBAgcdB=1
105 104 3ad2ant1 ABCA2+B2=C2¬2AAgcdBAAgcdBgcdBAgcdB=1
106 simp3 ABCA2+B2=C2¬2AAgcdB¬2AAgcdB
107 pythagtriplem18 AAgcdBBAgcdBCAgcdBAAgcdB2+BAgcdB2=CAgcdB2AAgcdBgcdBAgcdB=1¬2AAgcdBnmAAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2
108 25 38 77 100 105 106 107 syl312anc ABCA2+B2=C2¬2AAgcdBnmAAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2
109 93 86 11 divcan2d ABCAgcdBAAgcdB=A
110 109 eqcomd ABCA=AgcdBAAgcdB
111 96 86 11 divcan2d ABCAgcdBBAgcdB=B
112 111 eqcomd ABCB=AgcdBBAgcdB
113 85 86 11 divcan2d ABCAgcdBCAgcdB=C
114 113 eqcomd ABCC=AgcdBCAgcdB
115 110 112 114 3jca ABCA=AgcdBAAgcdBB=AgcdBBAgcdBC=AgcdBCAgcdB
116 115 3ad2ant1 ABCA2+B2=C2¬2AAgcdBA=AgcdBAAgcdBB=AgcdBBAgcdBC=AgcdBCAgcdB
117 oveq2 AAgcdB=m2n2AgcdBAAgcdB=AgcdBm2n2
118 117 eqeq2d AAgcdB=m2n2A=AgcdBAAgcdBA=AgcdBm2n2
119 118 3ad2ant1 AAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2A=AgcdBAAgcdBA=AgcdBm2n2
120 oveq2 BAgcdB=2mnAgcdBBAgcdB=AgcdB2mn
121 120 eqeq2d BAgcdB=2mnB=AgcdBBAgcdBB=AgcdB2mn
122 121 3ad2ant2 AAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2B=AgcdBBAgcdBB=AgcdB2mn
123 oveq2 CAgcdB=m2+n2AgcdBCAgcdB=AgcdBm2+n2
124 123 eqeq2d CAgcdB=m2+n2C=AgcdBCAgcdBC=AgcdBm2+n2
125 124 3ad2ant3 AAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2C=AgcdBCAgcdBC=AgcdBm2+n2
126 119 122 125 3anbi123d AAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2A=AgcdBAAgcdBB=AgcdBBAgcdBC=AgcdBCAgcdBA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
127 116 126 syl5ibcom ABCA2+B2=C2¬2AAgcdBAAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2A=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
128 127 reximdv ABCA2+B2=C2¬2AAgcdBmAAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2mA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
129 128 reximdv ABCA2+B2=C2¬2AAgcdBnmAAgcdB=m2n2BAgcdB=2mnCAgcdB=m2+n2nmA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
130 108 129 mpd ABCA2+B2=C2¬2AAgcdBnmA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
131 oveq1 k=AgcdBkm2n2=AgcdBm2n2
132 131 eqeq2d k=AgcdBA=km2n2A=AgcdBm2n2
133 oveq1 k=AgcdBk2mn=AgcdB2mn
134 133 eqeq2d k=AgcdBB=k2mnB=AgcdB2mn
135 oveq1 k=AgcdBkm2+n2=AgcdBm2+n2
136 135 eqeq2d k=AgcdBC=km2+n2C=AgcdBm2+n2
137 132 134 136 3anbi123d k=AgcdBA=km2n2B=k2mnC=km2+n2A=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
138 137 2rexbidv k=AgcdBnmA=km2n2B=k2mnC=km2+n2nmA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2
139 138 rspcev AgcdBnmA=AgcdBm2n2B=AgcdB2mnC=AgcdBm2+n2knmA=km2n2B=k2mnC=km2+n2
140 3 130 139 syl2anc ABCA2+B2=C2¬2AAgcdBknmA=km2n2B=k2mnC=km2+n2
141 rexcom knmA=km2n2B=k2mnC=km2+n2nkmA=km2n2B=k2mnC=km2+n2
142 rexcom kmA=km2n2B=k2mnC=km2+n2mkA=km2n2B=k2mnC=km2+n2
143 142 rexbii nkmA=km2n2B=k2mnC=km2+n2nmkA=km2n2B=k2mnC=km2+n2
144 141 143 bitri knmA=km2n2B=k2mnC=km2+n2nmkA=km2n2B=k2mnC=km2+n2
145 140 144 sylib ABCA2+B2=C2¬2AAgcdBnmkA=km2n2B=k2mnC=km2+n2