Metamath Proof Explorer


Theorem pythagtriplem13

Description: Lemma for pythagtrip . Show that N (which will eventually be closely related to the n in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Hypothesis pythagtriplem13.1 N=C+BCB2
Assertion pythagtriplem13 ABCA2+B2=C2AgcdB=1¬2AN

Proof

Step Hyp Ref Expression
1 pythagtriplem13.1 N=C+BCB2
2 pythagtriplem9 ABCA2+B2=C2AgcdB=1¬2AC+B
3 2 nnzd ABCA2+B2=C2AgcdB=1¬2AC+B
4 simp3r ABCA2+B2=C2AgcdB=1¬2A¬2A
5 2z 2
6 simp3 ABCC
7 simp2 ABCB
8 6 7 nnaddcld ABCC+B
9 8 nnzd ABCC+B
10 9 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2AC+B
11 nnz AA
12 11 3ad2ant1 ABCA
13 12 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2AA
14 dvdsgcdb 2C+BA2C+B2A2C+BgcdA
15 5 10 13 14 mp3an2i ABCA2+B2=C2AgcdB=1¬2A2C+B2A2C+BgcdA
16 15 biimpar ABCA2+B2=C2AgcdB=1¬2A2C+BgcdA2C+B2A
17 16 simprd ABCA2+B2=C2AgcdB=1¬2A2C+BgcdA2A
18 4 17 mtand ABCA2+B2=C2AgcdB=1¬2A¬2C+BgcdA
19 pythagtriplem7 ABCA2+B2=C2AgcdB=1¬2AC+B=C+BgcdA
20 19 breq2d ABCA2+B2=C2AgcdB=1¬2A2C+B2C+BgcdA
21 18 20 mtbird ABCA2+B2=C2AgcdB=1¬2A¬2C+B
22 pythagtriplem8 ABCA2+B2=C2AgcdB=1¬2ACB
23 22 nnzd ABCA2+B2=C2AgcdB=1¬2ACB
24 nnz CC
25 24 3ad2ant3 ABCC
26 nnz BB
27 26 3ad2ant2 ABCB
28 25 27 zsubcld ABCCB
29 28 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2ACB
30 dvdsgcdb 2CBA2CB2A2CBgcdA
31 5 29 13 30 mp3an2i ABCA2+B2=C2AgcdB=1¬2A2CB2A2CBgcdA
32 31 biimpar ABCA2+B2=C2AgcdB=1¬2A2CBgcdA2CB2A
33 32 simprd ABCA2+B2=C2AgcdB=1¬2A2CBgcdA2A
34 4 33 mtand ABCA2+B2=C2AgcdB=1¬2A¬2CBgcdA
35 pythagtriplem6 ABCA2+B2=C2AgcdB=1¬2ACB=CBgcdA
36 35 breq2d ABCA2+B2=C2AgcdB=1¬2A2CB2CBgcdA
37 34 36 mtbird ABCA2+B2=C2AgcdB=1¬2A¬2CB
38 omoe C+B¬2C+BCB¬2CB2C+BCB
39 3 21 23 37 38 syl22anc ABCA2+B2=C2AgcdB=1¬2A2C+BCB
40 28 zred ABCCB
41 40 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2ACB
42 simp13 ABCA2+B2=C2AgcdB=1¬2AC
43 42 nnred ABCA2+B2=C2AgcdB=1¬2AC
44 8 nnred ABCC+B
45 44 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2AC+B
46 nnrp BB+
47 46 3ad2ant2 ABCB+
48 47 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2AB+
49 43 48 ltsubrpd ABCA2+B2=C2AgcdB=1¬2ACB<C
50 nngt0 B0<B
51 50 3ad2ant2 ABC0<B
52 51 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2A0<B
53 simp12 ABCA2+B2=C2AgcdB=1¬2AB
54 53 nnred ABCA2+B2=C2AgcdB=1¬2AB
55 54 43 ltaddposd ABCA2+B2=C2AgcdB=1¬2A0<BC<C+B
56 52 55 mpbid ABCA2+B2=C2AgcdB=1¬2AC<C+B
57 41 43 45 49 56 lttrd ABCA2+B2=C2AgcdB=1¬2ACB<C+B
58 pythagtriplem10 ABCA2+B2=C20<CB
59 58 3adant3 ABCA2+B2=C2AgcdB=1¬2A0<CB
60 0re 0
61 ltle 0CB0<CB0CB
62 60 61 mpan CB0<CB0CB
63 41 59 62 sylc ABCA2+B2=C2AgcdB=1¬2A0CB
64 nngt0 C0<C
65 64 3ad2ant3 ABC0<C
66 65 3ad2ant1 ABCA2+B2=C2AgcdB=1¬2A0<C
67 43 54 66 52 addgt0d ABCA2+B2=C2AgcdB=1¬2A0<C+B
68 ltle 0C+B0<C+B0C+B
69 60 68 mpan C+B0<C+B0C+B
70 45 67 69 sylc ABCA2+B2=C2AgcdB=1¬2A0C+B
71 41 63 45 70 sqrtltd ABCA2+B2=C2AgcdB=1¬2ACB<C+BCB<C+B
72 57 71 mpbid ABCA2+B2=C2AgcdB=1¬2ACB<C+B
73 nnsub CBC+BCB<C+BC+BCB
74 22 2 73 syl2anc ABCA2+B2=C2AgcdB=1¬2ACB<C+BC+BCB
75 72 74 mpbid ABCA2+B2=C2AgcdB=1¬2AC+BCB
76 75 nnzd ABCA2+B2=C2AgcdB=1¬2AC+BCB
77 evend2 C+BCB2C+BCBC+BCB2
78 76 77 syl ABCA2+B2=C2AgcdB=1¬2A2C+BCBC+BCB2
79 39 78 mpbid ABCA2+B2=C2AgcdB=1¬2AC+BCB2
80 75 nngt0d ABCA2+B2=C2AgcdB=1¬2A0<C+BCB
81 75 nnred ABCA2+B2=C2AgcdB=1¬2AC+BCB
82 halfpos2 C+BCB0<C+BCB0<C+BCB2
83 81 82 syl ABCA2+B2=C2AgcdB=1¬2A0<C+BCB0<C+BCB2
84 80 83 mpbid ABCA2+B2=C2AgcdB=1¬2A0<C+BCB2
85 elnnz C+BCB2C+BCB20<C+BCB2
86 79 84 85 sylanbrc ABCA2+B2=C2AgcdB=1¬2AC+BCB2
87 1 86 eqeltrid ABCA2+B2=C2AgcdB=1¬2AN