Metamath Proof Explorer


Theorem pythagtriplem11

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

Ref Expression
Hypothesis pythagtriplem11.1 M = C + B + C B 2
Assertion pythagtriplem11 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A M

Proof

Step Hyp Ref Expression
1 pythagtriplem11.1 M = C + B + C B 2
2 pythagtriplem9 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B
3 2 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B
4 simp3r A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A ¬ 2 A
5 2z 2
6 nnz C C
7 6 3ad2ant3 A B C C
8 nnz B B
9 8 3ad2ant2 A B C B
10 7 9 zaddcld A B C C + B
11 10 3ad2ant1 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B
12 nnz A A
13 12 3ad2ant1 A B C A
14 13 3ad2ant1 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A A
15 dvdsgcdb 2 C + B A 2 C + B 2 A 2 C + B gcd A
16 5 11 14 15 mp3an2i A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B 2 A 2 C + B gcd A
17 16 biimpar A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B gcd A 2 C + B 2 A
18 17 simprd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B gcd A 2 A
19 4 18 mtand A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A ¬ 2 C + B gcd A
20 pythagtriplem7 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B = C + B gcd A
21 20 breq2d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B 2 C + B gcd A
22 19 21 mtbird A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A ¬ 2 C + B
23 pythagtriplem8 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C B
24 23 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C B
25 7 9 zsubcld A B C C B
26 25 3ad2ant1 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C B
27 dvdsgcdb 2 C B A 2 C B 2 A 2 C B gcd A
28 5 26 14 27 mp3an2i A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C B 2 A 2 C B gcd A
29 28 biimpar A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C B gcd A 2 C B 2 A
30 29 simprd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C B gcd A 2 A
31 4 30 mtand A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A ¬ 2 C B gcd A
32 pythagtriplem6 A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C B = C B gcd A
33 32 breq2d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C B 2 C B gcd A
34 31 33 mtbird A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A ¬ 2 C B
35 opoe C + B ¬ 2 C + B C B ¬ 2 C B 2 C + B + C B
36 3 22 24 34 35 syl22anc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B + C B
37 2 23 nnaddcld A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B + C B
38 37 nnzd A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B + C B
39 evend2 C + B + C B 2 C + B + C B C + B + C B 2
40 38 39 syl A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 2 C + B + C B C + B + C B 2
41 36 40 mpbid A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B + C B 2
42 2 nnred A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B
43 23 nnred A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C B
44 2 nngt0d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 0 < C + B
45 23 nngt0d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 0 < C B
46 42 43 44 45 addgt0d A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 0 < C + B + C B
47 37 nnred A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B + C B
48 halfpos2 C + B + C B 0 < C + B + C B 0 < C + B + C B 2
49 47 48 syl A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 0 < C + B + C B 0 < C + B + C B 2
50 46 49 mpbid A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A 0 < C + B + C B 2
51 elnnz C + B + C B 2 C + B + C B 2 0 < C + B + C B 2
52 41 50 51 sylanbrc A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A C + B + C B 2
53 1 52 eqeltrid A B C A 2 + B 2 = C 2 A gcd B = 1 ¬ 2 A M